Sau đây là danh sách các tích phân (nguyên hàm) của các hàm phân thức. Tích phân của mọi hàm phân thức đều có thể được tính bằng phân tích phân số một phần thành tổng các hàm số có dạng:
, và ![{\displaystyle {\frac {ax+b}{\left((x-c)^{2}+d^{2}\right)^{n}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/277e9eff8e906c97a0ea9b8d3dcf69566a5aa4d7)
rồi lần lượt xử lý từng số hạng.
Với những dạng hàm số khác, xem danh sách tích phân.
Hàm có dạng xm(ax + b)n
Nhiều nguyên hàm dưới đây có hạng tử dạng ln . Do hạng tử này không có nghĩa khi x = −b / a, dạng tổng quát của nguyên hàm thay hằng số tích phân bằng một hàm hằng cục bộ.[1] Tuy nhiên, người ta thường bỏ nó ra khỏi biểu thức. Ví dụ
![{\displaystyle \int {\frac {1}{ax+b}}\,dx={\begin{cases}{\dfrac {1}{a}}\ln(-(ax+b))+C^{-}&ax+b<0\\{\dfrac {1}{a}}\ln(ax+b)+C^{+}&ax+b>0\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23c6fcc9f6ed4c40dbb0c816510eb6b56c4f2689)
thường được viết ngắn gọn là
![{\displaystyle \int {\frac {1}{ax+b}}\,dx={\frac {1}{a}}\ln \left|ax+b\right|+C,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39dbfed3cadfe902b99fe8d1450a31a84b8ecf5a)
trong đó C được hiểu là ký hiệu cho hàm hằng cục bộ ẩn x. Quy ước này sẽ được tuân theo trong phần còn lại This convention will be adhered to in the following.
(Công thức diện tích Cavalieri) ![{\displaystyle \int {\frac {x}{ax+b}}\,dx={\frac {x}{a}}-{\frac {b}{a^{2}}}\ln \left|ax+b\right|+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0044cf33077ed2838244a1f49755bfda52d1b5c2)
![{\displaystyle \int {\frac {x}{(ax+b)^{2}}}\,dx={\frac {b}{a^{2}(ax+b)}}+{\frac {1}{a^{2}}}\ln \left|ax+b\right|+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fca8db2b5e2c57c5991f42012f3fe019e82d6db8)
![{\displaystyle \int {\frac {x}{(ax+b)^{n}}}\,dx={\frac {a(1-n)x-b}{a^{2}(n-1)(n-2)(ax+b)^{n-1}}}+C\qquad (n\not \in \{1,2\}{\mbox{)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a9e5e03388e52f6f86942da2663c353744addb0)
![{\displaystyle \int x(ax+b)^{n}\,dx={\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)}}(ax+b)^{n+1}+C\qquad (n\not \in \{-1,-2\}{\mbox{)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2628e131a7705b94b2a070e06a9b6b7c7c475751)
![{\displaystyle \int {\frac {x^{2}}{ax+b}}\,dx={\frac {b^{2}\ln(\left|ax+b\right|)}{a^{3}}}+{\frac {ax^{2}-2bx}{2a^{2}}}+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52efc74406ece758b265c368e08dd9ee16c8a8fc)
![{\displaystyle \int {\frac {x^{2}}{(ax+b)^{2}}}\,dx={\frac {1}{a^{3}}}\left(ax-2b\ln \left|ax+b\right|-{\frac {b^{2}}{ax+b}}\right)+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9bbcfb06f106d8fe208d68fc69d74c426a18ce8c)
![{\displaystyle \int {\frac {x^{2}}{(ax+b)^{3}}}\,dx={\frac {1}{a^{3}}}\left(\ln \left|ax+b\right|+{\frac {2b}{ax+b}}-{\frac {b^{2}}{2(ax+b)^{2}}}\right)+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8d9738490df1b57b94860a05b22b0ff7c61237d)
![{\displaystyle \int {\frac {x^{2}}{(ax+b)^{n}}}\,dx={\frac {1}{a^{3}}}\left(-{\frac {(ax+b)^{3-n}}{(n-3)}}+{\frac {2b(ax+b)^{2-n}}{(n-2)}}-{\frac {b^{2}(ax+b)^{1-n}}{(n-1)}}\right)+C\qquad (n\not \in \{1,2,3\}{\mbox{)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7093270329930b242e067c04f69e4e00efb7483)
![{\displaystyle \int {\frac {1}{x(ax+b)}}\,dx=-{\frac {1}{b}}\ln \left|{\frac {ax+b}{x}}\right|+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0eb20d2f8f53fa28110b76a629615131429bbdea)
![{\displaystyle \int {\frac {1}{x^{2}(ax+b)}}\,dx=-{\frac {1}{bx}}+{\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d446865683df020ec83ce8838cdcae2f1d2bd5f)
![{\displaystyle \int {\frac {1}{x^{2}(ax+b)^{2}}}\,dx=-a\left({\frac {1}{b^{2}(ax+b)}}+{\frac {1}{ab^{2}x}}-{\frac {2}{b^{3}}}\ln \left|{\frac {ax+b}{x}}\right|\right)+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e03e60c23681f5186b4a95a5e9ea921732e122a)
Hàm có dạng xm / (a x2 + b x + c)n
Với a ≠ 0:
![{\displaystyle \int {\frac {1}{ax^{2}+bx+c}}dx={\begin{cases}\displaystyle {\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C&(4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle {\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C={\begin{cases}\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac}}}\,\operatorname {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&(|2ax+b|<{\sqrt {b^{2}-4ac}}{\mbox{)}}\\[6pt]\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac}}}\,\operatorname {arccoth} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&(|2ax+b|\geq {\sqrt {b^{2}-4ac}}{\mbox{)}}\end{cases}}&(4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle -{\frac {2}{2ax+b}}+C&(4ac-b^{2}=0{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e7a5fadcc2b04bbac9652cdbde06df445395fd5)
![{\displaystyle \int {\frac {x}{ax^{2}+bx+c}}\,dx={\frac {1}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {b}{2a}}\int {\frac {dx}{ax^{2}+bx+c}}+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a2741cb28c134c0cfe27162393974e3c56475461)
![{\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}\,dx={\begin{cases}\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {4ac-b^{2}}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C&(4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\operatorname {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&(4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}}+C&(4ac-b^{2}=0{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40a221896ecd027b29e594b8ffc71c7e1a0ce83d)
![{\displaystyle \int {\frac {1}{(ax^{2}+bx+c)^{n}}}\,dx={\frac {2ax+b}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}+{\frac {(2n-3)2a}{(n-1)(4ac-b^{2})}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}\,dx+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/70aa274c4d40db21d09cced28142d167f4fd1aab)
![{\displaystyle \int {\frac {x}{(ax^{2}+bx+c)^{n}}}\,dx=-{\frac {bx+2c}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}-{\frac {b(2n-3)}{(n-1)(4ac-b^{2})}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}\,dx+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9596eeb7ecf42279db0fe8ddbff66b7aa5ffca3)
![{\displaystyle \int {\frac {1}{x(ax^{2}+bx+c)}}\,dx={\frac {1}{2c}}\ln \left|{\frac {x^{2}}{ax^{2}+bx+c}}\right|-{\frac {b}{2c}}\int {\frac {1}{ax^{2}+bx+c}}\,dx+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce3a0fffb5a4379f6c63480665bb09f9d6d1f61b)
Hàm có dạng xm (a + b xn)p
- Những công thức sau hạ số mũ của hàm dưới dấu tích phân nhưng vẫn giữ nguyên dạng của chúng, do đó có thể được dùng nhiều lần để đưa số mũ m và p xuống 0.
- Những công thức hạ bậc này có thể dùng cho hàm có số mũ nguyên hoặc hữu tỉ.
![{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p}}{m+n\,p+1}}\,+\,{\frac {a\,n\,p}{m+n\,p+1}}\int x^{m}\left(a+b\,x^{n}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4790301459e72262aafa18b525a4f5e3bdf47c07)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx=-{\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p+1}}{a\,n(p+1)}}\,+\,{\frac {m+n(p+1)+1}{a\,n(p+1)}}\int x^{m}\left(a+b\,x^{n}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e2c8816cb99165f737c15e4d99c66d493f720137)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p}}{m+1}}\,-\,{\frac {b\,n\,p}{m+1}}\int x^{m+n}\left(a+b\,x^{n}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96bba069ebcdb45e98e0062bb4b07008b0d35fc2)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}}{b\,n(p+1)}}\,-\,{\frac {m-n+1}{b\,n(p+1)}}\int x^{m-n}\left(a+b\,x^{n}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3422de38d52f4da42c1bf9968ce5768120c7086)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}}{b(m+n\,p+1)}}\,-\,{\frac {a(m-n+1)}{b(m+n\,p+1)}}\int x^{m-n}\left(a+b\,x^{n}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26dfd412da178d54770054a6acc2eb5bfe05fa4e)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p+1}}{a(m+1)}}\,-\,{\frac {b(m+n(p+1)+1)}{a(m+1)}}\int x^{m+n}\left(a+b\,x^{n}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/853386481e1c31cde5b48c5df4f22b78d7b39324)
Hàm có dạng (A + B x) (a + b x)m (c + d x)n (e + f x)p
- Tương tự như trên, những công thức hạ bậc này có thể được dùng nhiều lần để đưa m, n và p xuống 0.
- Những công thức này dùng được cho số mũ là số nguyên hoặc số hữu tỉ.
- Cho B bằng 0, ta có trường hợp đặc biệt
.
![{\displaystyle \int (A+B\,x)(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}dx=-{\frac {(A\,b-a\,B)(a+b\,x)^{m+1}(c+d\,x)^{n}(e+f\,x)^{p+1}}{b(m+1)(a\,f-b\,e)}}\,+\,{\frac {1}{b(m+1)(a\,f-b\,e)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c987f05158ae24bae440d927972e8a63211243f5)
![{\displaystyle \int (b\,c(m+1)(A\,f-B\,e)+(A\,b-a\,B)(n\,d\,e+c\,f(p+1))+d(b(m+1)(A\,f-B\,e)+f(n+p+1)(A\,b-a\,B))x)(a+b\,x)^{m+1}(c+d\,x)^{n-1}(e+f\,x)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0b28e50be43852e6d20bf28ee71196851cb153e)
![{\displaystyle \int (A+B\,x)(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}dx={\frac {B(a+b\,x)^{m}(c+d\,x)^{n+1}(e+f\,x)^{p+1}}{d\,f(m+n+p+2)}}\,+\,{\frac {1}{d\,f(m+n+p+2)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/aab7fb9f464b056e6fc58cd059c10a37c515b9ee)
![{\displaystyle \int (A\,a\,d\,f(m+n+p+2)-B(b\,c\,e\,m+a(d\,e(n+1)+c\,f(p+1)))+(A\,b\,d\,f(m+n+p+2)+B(a\,d\,f\,m-b(d\,e(m+n+1)+c\,f(m+p+1))))x)(a+b\,x)^{m-1}(c+d\,x)^{n}(e+f\,x)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a655aa2db92c27864331d30368cb0adf8c70a56)
![{\displaystyle \int (A+B\,x)(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}dx={\frac {(A\,b-a\,B)(a+b\,x)^{m+1}(c+d\,x)^{n+1}(e+f\,x)^{p+1}}{(m+1)(a\,d-b\,c)(a\,f-b\,e)}}\,+\,{\frac {1}{(m+1)(a\,d-b\,c)(a\,f-b\,e)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/18dba981fd20c81e32ae51178b701cdd5164017a)
![{\displaystyle \int ((m+1)(A(a\,d\,f-b(c\,f+d\,e))+B\,b\,c\,e)-(A\,b-a\,B)(d\,e(n+1)+c\,f(p+1))-d\,f(m+n+p+3)(A\,b-a\,B)x)(a+b\,x)^{m+1}(c+d\,x)^{n}(e+f\,x)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdb89a9f5d6c5f99bb63971b5606c14cddb2820d)
Hàm có dạng xm (A + B xn) (a + b xn)p (c + d xn)q
![{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx=-{\frac {(A\,b-a\,B)x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}}{a\,b\,n(p+1)}}\,+\,{\frac {1}{a\,b\,n(p+1)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a38437da40d4bfb53926163ab43e307e218ee1ab)
![{\displaystyle \int x^{m}\left(c(A\,b\,n(p+1)+(A\,b-a\,B)(m+1))+d(A\,b\,n(p+1)+(A\,b-a\,B)(m+n\,q+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c0439c37e0d84e040a7d386104f14a1918eb5f7)
![{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {B\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}}{b(m+n(p+q+1)+1)}}\,+\,{\frac {1}{b(m+n(p+q+1)+1)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1345a3383a8a06d2b9e927c2bb1164c3882dcc6c)
![{\displaystyle \int x^{m}\left(c((A\,b-a\,B)(1+m)+A\,b\,n(1+p+q))+(d(A\,b-a\,B)(1+m)+B\,n\,q(b\,c-a\,d)+A\,b\,d\,n(1+p+q))\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7727bbca249e0388c32fcc19ab2fe79f01264390)
![{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx=-{\frac {(A\,b-a\,B)x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{a\,n(b\,c-a\,d)(p+1)}}\,+\,{\frac {1}{a\,n(b\,c-a\,d)(p+1)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c5040089a6b24b57adbe22a41f38bf24afb4b8f)
![{\displaystyle \int x^{m}\left(c(A\,b-a\,B)(m+1)+A\,n(b\,c-a\,d)(p+1)+d(A\,b-a\,B)(m+n(p+q+2)+1)x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c315f6222e8402c188ec8f2fddd4d0b2b99f1e5)
![{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {B\,x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{b\,d(m+n(p+q+1)+1)}}\,-\,{\frac {1}{b\,d(m+n(p+q+1)+1)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ed26f4e23264862d796a95aea8545ac15fe1ec0)
![{\displaystyle \int x^{m-n}\left(a\,B\,c(m-n+1)+(a\,B\,d(m+n\,q+1)-b(-B\,c(m+n\,p+1)+A\,d(m+n(p+q+1)+1)))x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d8f1f0b4716f44dafdba65b02c8bd869a21e0f4)
![{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {A\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{a\,c(m+1)}}\,+\,{\frac {1}{a\,c(m+1)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/10cc116a239a3b3f734ff82656873cc92f1eaaf6)
![{\displaystyle \int x^{m+n}\left(a\,B\,c(m+1)-A(b\,c+a\,d)(m+n+1)-A\,n(b\,c\,p+a\,d\,q)-A\,b\,d(m+n(p+q+2)+1)x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f12372d1fc23dc99e0c3fea80274856ddd31c94)
![{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {A\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}}{a(m+1)}}\,-\,{\frac {1}{a(m+1)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/91818b70c6c337a8c33e5a108d8784a0af1058c7)
![{\displaystyle \int x^{m+n}\left(c(A\,b-a\,B)(m+1)+A\,n(b\,c(p+1)+a\,d\,q)+d((A\,b-a\,B)(m+1)+A\,b\,n(p+q+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e31fe173375ce97b040fe8e83d767f86544f15e)
![{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {(A\,b-a\,B)x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{b\,n(b\,c-a\,d)(p+1)}}\,-\,{\frac {1}{b\,n(b\,c-a\,d)(p+1)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2967fbd56728834ec1b55db584498f7d77d57354)
![{\displaystyle \int x^{m-n}\left(c(A\,b-a\,B)(m-n+1)+(d(A\,b-a\,B)(m+n\,q+1)-b\,n(B\,c-A\,d)(p+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c4eebe06b1b7e2f5ad7b4532eded00f124ac4b9)
Hàm có dạng (d + e x)m (a + b x + c x2)p với b2 − 4 a c = 0
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p}}{e(m+1)}}\,-\,{\frac {p(d+e\,x)^{m+2}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p-1}}{e^{2}(m+1)(m+2p+1)}}\,+\,{\frac {p(2p-1)(2c\,d-b\,e)}{e^{2}(m+1)(m+2p+1)}}\int (d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81d4c261f463293ee05481519bb1634bf765009b)
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p}}{e(m+1)}}\,-\,{\frac {p(d+e\,x)^{m+2}(b+2\,c\,x)\left(a+b\,x+c\,x^{2}\right)^{p-1}}{e^{2}(m+1)(m+2)}}\,+\,{\frac {2\,c\,p\,(2\,p-1)}{e^{2}(m+1)(m+2)}}\int (d+e\,x)^{m+2}\left(a+b\,x+c\,x^{2}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52be3971d00f1b06cbb659d213cb6e875c7beb23)
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {e(m+2p+2)(d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p+1}}{(p+1)(2p+1)(2c\,d-b\,e)}}\,+\,{\frac {(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p}}{(2p+1)(2c\,d-b\,e)}}\,+\,{\frac {e^{2}m(m+2p+2)}{(p+1)(2p+1)(2c\,d-b\,e)}}\int (d+e\,x)^{m-1}\left(a+b\,x+c\,x^{2}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd7c90e7a6da4a559c53a66731a2d51e3533e7af)
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {e\,m(d+e\,x)^{m-1}\left(a+b\,x+c\,x^{2}\right)^{p+1}}{2c(p+1)(2p+1)}}\,+\,{\frac {(d+e\,x)^{m}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p}}{2c(2p+1)}}\,+\,{\frac {e^{2}m(m-1)}{2c(p+1)(2p+1)}}\int (d+e\,x)^{m-2}\left(a+b\,x+c\,x^{2}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48f0d694e4cffcb65184a89244a10abaf0b9b346)
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p}}{e(m+2p+1)}}\,-\,{\frac {p(2c\,d-b\,e)(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p-1}}{2c\,e^{2}(m+2p)(m+2p+1)}}\,+\,{\frac {p(2p-1)(2c\,d-b\,e)^{2}}{2c\,e^{2}(m+2p)(m+2p+1)}}\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5639f373ae3a1cbf3900954d75e024c850b57be5)
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {2c\,e(m+2p+2)(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p+1}}{(p+1)(2p+1)(2c\,d-b\,e)^{2}}}\,+\,{\frac {(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p}}{(2p+1)(2c\,d-b\,e)}}\,+\,{\frac {2c\,e^{2}(m+2p+2)(m+2p+3)}{(p+1)(2p+1)(2c\,d-b\,e)^{2}}}\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5cfc86b744b2f318e9e0a51ec13b7576e7b53917)
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p}}{2c(m+2p+1)}}\,+\,{\frac {m(2c\,d-b\,e)}{2c(m+2p+1)}}\int (d+e\,x)^{m-1}\left(a+b\,x+c\,x^{2}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9a0ad7c5a71f6bcb324373ed772deb8ec0a6e1a)
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p}}{(m+1)(2c\,d-b\,e)}}\,+\,{\frac {2c(m+2p+2)}{(m+1)(2c\,d-b\,e)}}\int (d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f81c419c1332e643d09afb2f927a42c10c468bb)
Hàm có dạng (d + e x)m (A + B x) (a + b x + c x2)p
![{\displaystyle \int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}(A\,e(m+2p+2)-B\,d(2p+1)+e\,B(m+1)x)\left(a+b\,x+c\,x^{2}\right)^{p}}{e^{2}(m+1)(m+2p+2)}}\,+\,{\frac {1}{e^{2}(m+1)(m+2p+2)}}p\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/78697ac4e74dffc04fc78d19dd9087db122a3346)
![{\displaystyle \int (d+e\,x)^{m+1}(B(b\,d+2a\,e+2a\,e\,m+2b\,d\,p)-A\,b\,e(m+2p+2)+(B(2c\,d+b\,e+b\,em+4c\,d\,p)-2A\,c\,e(m+2p+2))x)\left(a+b\,x+c\,x^{2}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb73a7dec47f797fd9278205be008dd9bc663e49)
![{\displaystyle \int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m}(A\,b-2a\,B-(b\,B-2A\,c)x)\left(a+b\,x+c\,x^{2}\right)^{p+1}}{(p+1)\left(b^{2}-4a\,c\right)}}\,+\,{\frac {1}{(p+1)\left(b^{2}-4a\,c\right)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/83f35ba16e837d182a6ed5fef53717fd25ac27d9)
![{\displaystyle \int (d+e\,x)^{m-1}(B(2a\,e\,m+b\,d(2p+3))-A(b\,e\,m+2c\,d(2p+3))+e(b\,B-2A\,c)(m+2p+3)x)\left(a+b\,x+c\,x^{2}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d7df1bf0df1e3904b2cf67eec6130d1c6c64fe4)
![{\displaystyle \int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}(A\,c\,e(m+2p+2)-B(c\,d+2c\,d\,p-b\,e\,p)+B\,c\,e(m+2p+1)x)\left(a+b\,x+c\,x^{2}\right)^{p}}{c\,e^{2}(m+2p+1)(m+2p+2)}}\,-\,{\frac {p}{c\,e^{2}(m+2p+1)(m+2p+2)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/17ddd861aa65035abdb6fb4c4b2f3aaa2a8be49d)
![{\displaystyle \int (d+e\,x)^{m}(A\,c\,e(b\,d-2a\,e)(m+2p+2)+B(a\,e(b\,e-2c\,d\,m+b\,e\,m)+b\,d(b\,e\,p-c\,d-2c\,d\,p))+}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98807707c17967e647b352809bc43126951a3d86)
![{\displaystyle \left(A\,c\,e(2c\,d-b\,e)(m+2p+2)-B\left(-b^{2}e^{2}(m+p+1)+2c^{2}d^{2}(1+2p)+c\,e(b\,d(m-2p)+2a\,e(m+2p+1))\right)\right)x)\left(a+b\,x+c\,x^{2}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f14c396c0e03ac3ba23d09f2590da9014d575600)
![{\displaystyle \int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(A\left(b\,c\,d-b^{2}e+2a\,c\,e\right)-a\,B(2c\,d-b\,e)+c(A(2c\,d-b\,e)-B(b\,d-2a\,e))x\right)\left(a+b\,x+c\,x^{2}\right)^{p+1}}{(p+1)\left(b^{2}-4a\,c\right)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)}}\,+}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e669a50034773bdb11361d29a1bebddaf9420901)
![{\displaystyle {\frac {1}{(p+1)\left(b^{2}-4a\,c\right)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/016e4c6aa45b533aa68585ddb811fc15308a748c)
![{\displaystyle \int (d+e\,x)^{m}(A\left(b\,c\,d\,e(2p-m+2)+b^{2}e^{2}(m+p+2)-2c^{2}d^{2}(3+2p)-2a\,c\,e^{2}(m+2p+3)\right)-}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b11e320462a5af2f18a06ef36ff6a376be06a6d)
![{\displaystyle B(a\,e(b\,e-2c\,dm+b\,e\,m)+b\,d(-3c\,d+b\,e-2c\,d\,p+b\,e\,p))+c\,e(B(b\,d-2a\,e)-A(2c\,d-b\,e))(m+2p+4)x)\left(a+b\,x+c\,x^{2}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5d23f0eaa894842f3ff9fabeda64bcf47fa94ec)
![{\displaystyle \int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {B(d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p+1}}{c(m+2p+2)}}\,+\,{\frac {1}{c(m+2p+2)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/57d105f18ed03f4ef773c0077e03d33d5f04951d)
![{\displaystyle \int (d+e\,x)^{m-1}(m(A\,c\,d-a\,B\,e)-d(b\,B-2A\,c)(p+1)+((B\,c\,d-b\,B\,e+A\,c\,e)m-e(b\,B-2A\,c)(p+1))x)\left(a+b\,x+c\,x^{2}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e87cefe15219284d13643c10f298de156804772)
![{\displaystyle \int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {(B\,d-A\,e)(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p+1}}{(m+1)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)}}\,+\,{\frac {1}{(m+1)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/afa56d4e2e2370939b2918e80223f6a68f4fb76f)
![{\displaystyle \int (d+e\,x)^{m+1}((A\,c\,d-A\,b\,e+a\,B\,e)(m+1)+b(B\,d-A\,e)(p+1)+c(B\,d-A\,e)(m+2p+3)x)\left(a+b\,x+c\,x^{2}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93e87350f7179d266357fcaca7919b2c8518cb1b)
Hàm có dạng xm (a + b xn + c x2n)p}} với {{math|1=b2 − 4 a c = 0
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{m+2n\,p+1}}\,+\,{\frac {n\,p\,x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}}{(m+1)(m+2n\,p+1)}}\,-\,{\frac {b\,n^{2}p(2p-1)}{(m+1)(m+2n\,p+1)}}\int x^{m+n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8b1f53bf9a1baa716e5af791b71798ab7c7409d)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {(m+n(2p-1)+1)x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{(m+1)(m+n+1)}}\,+\,{\frac {n\,p\,x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}}{(m+1)(m+n+1)}}\,+\,{\frac {2c\,p\,n^{2}(2p-1)}{(m+1)(m+n+1)}}\int x^{m+2n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/daf94843c8208eb4a740395f5c20467bd18f7890)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {(m+n(2p+1)+1)x^{m-n+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{b\,n^{2}(p+1)(2p+1)}}\,-\,{\frac {x^{m+1}\left(b+2c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{b\,n(2p+1)}}\,-\,{\frac {(m-n+1)(m+n(2p+1)+1)}{b\,n^{2}(p+1)(2p+1)}}\int x^{m-n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/021f1e8e9598be88cfa1607774fcee9206cf2e6c)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx=-{\frac {(m-3n-2n\,p+1)x^{m-2n+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{2c\,n^{2}(p+1)(2p+1)}}\,-\,{\frac {x^{m-2n+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{2c\,n(2p+1)}}\,+\,{\frac {(m-n+1)(m-2n+1)}{2c\,n^{2}(p+1)(2p+1)}}\int x^{m-2n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5519f9ce654c5b38b15c36e913ab34ed398b7a4e)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{m+2n\,p+1}}\,+\,{\frac {n\,p\,x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}}{(m+2n\,p+1)(m+n(2p-1)+1)}}\,+\,{\frac {2a\,n^{2}p(2p-1)}{(m+2n\,p+1)(m+n(2p-1)+1)}}\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f56db1ed0a596ae938014d966e70cec21caf6147)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx=-{\frac {(m+n+2n\,p+1)x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{2a\,n^{2}(p+1)(2p+1)}}\,-\,{\frac {x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{2a\,n(2p+1)}}\,+\,{\frac {(m+n(2p+1)+1)(m+2n(p+1)+1)}{2a\,n^{2}(p+1)(2p+1)}}\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38eababe79b99a4cd22d8a14cd9c524f14bd2c1d)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m-n+1}\left(b+2c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{2c(m+2n\,p+1)}}\,-\,{\frac {b(m-n+1)}{2c(m+2n\,p+1)}}\int x^{m-n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25d28b779eace5a71ad73fb2672f516197d71004)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(b+2c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{b(m+1)}}\,-\,{\frac {2c(m+n(2p+1)+1)}{b(m+1)}}\int x^{m+n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08f3bc0cc9c22df9b31c927fae0a91b906d41216)
Hàm có dạng xm (A + B xn) (a + b xn + c x2n)p
![{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(A(m+n(2p+1)+1)+B(m+1)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{(m+1)(m+n(2p+1)+1)}}\,+\,{\frac {n\,p}{(m+1)(m+n(2p+1)+1)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3facef4f9356ed7259e79b981b6d16fc339aae36)
![{\displaystyle \int x^{m+n}\left(2a\,B(m+1)-A\,b(m+n(2p+1)+1)+(b\,B(m+1)-2\,A\,c(m+n(2p+1)+1))x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d14e9827f511938766c163699b15f58747c5509b)
![{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m-n+1}\left(A\,b-2a\,B-(b\,B-2A\,c)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{n(p+1)\left(b^{2}-4a\,c\right)}}\,+\,{\frac {1}{n(p+1)\left(b^{2}-4a\,c\right)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e15dfb2d9c2d0465efe70f506957ea260fa29f9)
![{\displaystyle \int x^{m-n}\left((m-n+1)(2a\,B-A\,b)+(m+2n(p+1)+1)(b\,B-2A\,c)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ef33317b2f35c7e140ea19f003705c2c5f11bb4)
![{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(b\,B\,n\,p+A\,c(m+n(2p+1)+1)+B\,c(m+2n\,p+1)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{c(m+2n\,p+1)(m+n(2p+1)+1)}}\,+\,{\frac {n\,p}{c(m+2n\,p+1)(m+n(2p+1)+1)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/55ae5b543fd60eed6b9cfd5af052186543191e89)
![{\displaystyle \int x^{m}\left(2a\,A\,c(m+n(2p+1)+1)-a\,b\,B(m+1)+\left(2a\,B\,c(m+2n\,p+1)+A\,b\,c(m+n(2p+1)+1)-b^{2}B(m+n\,p+1)\right)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d003ed1b209497e70bcea5038c05e445c4441b9d)
![{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx=-{\frac {x^{m+1}\left(A\,b^{2}-a\,b\,B-2a\,A\,c+(A\,b-2a\,B)c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{a\,n(p+1)\left(b^{2}-4a\,c\right)}}\,+\,{\frac {1}{a\,n(p+1)\left(b^{2}-4a\,c\right)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/9978439d06b1bb98787ec28c833ac662344bc015)
![{\displaystyle \int x^{m}\left((m+n(p+1)+1)A\,b^{2}-a\,b\,B(m+1)-2(m+2n(p+1)+1)a\,A\,c+(m+n(2p+3)+1)(A\,b-2a\,B)c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc9c39004f70946f2d523d54eb5ac2a3e25a20de)
![{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {B\,x^{m-n+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{c(m+n(2p+1)+1)}}\,-\,{\frac {1}{c(m+n(2p+1)+1)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee44e9d56c64ded834d7f364a742b67f6dd8085f)
![{\displaystyle \int x^{m-n}\left(a\,B(m-n+1)+(b\,B(m+n\,p+1)-A\,c(m+n(2p+1)+1))x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d991f88480b32f7bf534eeb92821f25f5d35675)
![{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {A\,x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{a(m+1)}}\,+\,{\frac {1}{a(m+1)}}\,\cdot }](https://wikimedia.org/api/rest_v1/media/math/render/svg/67d07413686b8330b1ba3431fa974bdf11c26c5b)
![{\displaystyle \int x^{m+n}\left(a\,B(m+1)-A\,b(m+n(p+1)+1)-A\,c(m+2n(p+1)+1)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c2c74a020e5302e30ed495f96aea8694d275bdd)
Các hàm khác
![{\displaystyle \int {\frac {f'(x)}{f(x)}}\,dx=\ln \left|f(x)\right|+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68ff9dbfd7660c62e29ea8147600248c4ff07238)
![{\displaystyle \int {\frac {1}{x^{2}+a^{2}}}\,dx={\frac {1}{a}}\arctan {\frac {x}{a}}\,\!+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1acafa96f0cd8a2d9f05b62f34fef88a73225d3)
![{\displaystyle \int {\frac {1}{x^{2}-a^{2}}}\,dx={\begin{cases}\displaystyle -{\frac {1}{a}}\,\operatorname {arctanh} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}+C&(|x|<|a|{\mbox{)}}\\[12pt]\displaystyle -{\frac {1}{a}}\,\operatorname {arccoth} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}+C&(|x|>|a|{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36d2cad7041beb7ffb42553f5a30bfc4aa5dec29)
![{\displaystyle \int {\frac {dx}{x^{2^{n}}+1}}={\frac {1}{2^{n-1}}}\sum _{k=1}^{2^{n-1}}\sin \left({\frac {2k-1}{2^{n}}}\pi \right)\arctan \left[\left(x-\cos \left({\frac {2k-1}{2^{n}}}\pi \right)\right)\csc \left({\frac {2k-1}{2^{n}}}\pi \right)\right]-{\frac {1}{2}}\cos \left({\frac {2k-1}{2^{n}}}\pi \right)\ln \left|x^{2}-2x\cos \left({\frac {2k-1}{2^{n}}}\pi \right)+1\right|+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5086de865f6047c3f115bfbc3bd5ffde147a645c)
Tham khảo
- ^ "Reader Survey: log|x| + C", Tom Leinster, The n-category Café, 19 tháng 3, 2012