常函数、幂函数、指数函数
\[\int kdx = kx + C \, (k \in C) \]\[\int x^adx = \frac{x^{a + 1}}{a + 1} + C \, (a \neq -1) \]\[\int \frac{dx}{x} = \ln{\lvert x \rvert} + C \]\[\int a^xdx = \frac{a^x}{\ln{a}} + C \]\[\int e^xdx = e^x + C \]三角函数
\[\int\sin{x}dx = -\cos{x} + C \]\[\int\cos{x}dx = \sin{x} + C \]\[\int\sec^2{x}dx = \tan{x} + C \]\[\int\csc^2{x}dx = -\cot{x} + C \]\[\int\sec{x}\tan{x}dx = \sec{x} + C \]\[\int\csc{x}\cot{x}dx = -\csc{x} + C \]反三角函数
\[\int\frac{dx}{\sqrt{1 - x^2}} = \arcsin{x} + C \]\[\int-\frac{dx}{\sqrt{1 - x^2}} = \arccos{x} + C \]\[\int\frac{dx}{1 + x^2} = \arctan{x} + C \]\[\int-\frac{dx}{1 + x^2} = \operatorname{arccot}x + C \]双曲函数
\[\int\sinh{x} = \cosh{x} + C \]\[\int\cosh{x} = \sinh{x} + C \]复杂三角函数
\[\int\tan{x}dx = -\ln\lvert\cos{x}\rvert + C \]\[\int\cot{x}dx = \ln\lvert\sin{x}\rvert + C \]\[\int\sec{x}dx = \frac{1}{2}\ln\left\lvert\frac{\sin{x} + 1}{\sin{x} - 1}\right\rvert + C = \ln\lvert\sec{x} + \tan{x}\rvert + C \]\[\int\csc{x}dx = \frac{1}{2}\ln\left\lvert\frac{\cos{x} - 1}{\cos{x} +1}\right\rvert + C = \ln\lvert\csc{x} - \cot{x}\rvert + C = \ln\left\lvert\tan{\frac{x}{2}}\right\rvert + C \]复杂分式函数
\[\int\frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan{\frac{x}{a}} + C \]\[\int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left\lvert\frac{x - a}{x + a}\right\rvert + C \]\[\int\frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C \]\[\int\frac{dx}{\sqrt{x^2 + a^2}} = \ln\left(x + \sqrt{x^2 + a^2}\right) + C \]\[\int\frac{dx}{\sqrt{x^2 - a^2}} = \ln\left\lvert x + \sqrt{x^2 - a^2}\right\rvert + C \]部分积分公式的证明
一个关于三角函数你必须知道的半角公式：
\[\frac{\sin{x}}{1 + \cos{x}} = \frac{2\sin{\frac{x}{2}}\cos{\frac{x}{2}} }{1 + (2\cos^2{\frac{x}{2}} - 1)} = \tan\frac{x}{2} \]\[\frac{1 - \cos{x}}{\sin{x}} = \frac{1 - (1 - 2\sin^2{\frac{x}{2}})}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \tan\frac{x}{2} \]（1）\(\int\sec{x}dx = \frac{1}{2}\ln\left\lvert\frac{\sin{x} + 1}{\sin{x} - 1}\right\rvert + C = \ln\lvert\sec{x} + \tan{x}\rvert + C\)
对于第一个等号有：
\[\begin{aligned} \int\sec{x}dx &= \int\frac{cos{x}}{\cos^2{x}}dx \\ &= \int\frac{d(\sin{x})}{1 - \sin^2{x}} \\ &= \frac{1}{2}\left(\int\frac{d(\sin{x})}{1 - \sin{x}} - \int\frac{d(\sin{x})}{1 + \sin{x}}\right) \\ &= \frac{1}{2}\ln\left\lvert\frac{\sin{x} + 1}{\sin{x} - 1}\right\rvert + C \\ \end{aligned}\]由于不定积分 \(\int\sec^2{x}dx = \tan{x} + C\)，\(\int\sec{t}\tan{x}dx = \sec{x} + C\) ，原式可以转化为：
\[\begin{aligned} \int\sec{x}dx &= \int\frac{\sec{x}(\sec{x} + \tan{x})}{sec{x} + \tan{x}}dx \\ &= \int\frac{d(sec{x} + \tan{x})}{\sec{x} + \tan{x}} \\ &= \ln\lvert\sec{x} + \tan{x}\rvert + C \end{aligned}\]（2）\(\int\csc{x}dx = \frac{1}{2}\ln\left\lvert\frac{\cos{x} - 1}{\cos{x} +1}\right\rvert + C = \ln\lvert\csc{x} - \cot{x}\rvert + C = \ln\left\lvert\tan{\frac{x}{2}}\right\rvert + C\)
对于第一个等号有：
\[\begin{aligned} \int\csc{x}dx &= \int\frac{\sin{x}}{\sin^2{x}}dx \\ &= \int-\frac{d(\cos{x})}{1 - \cos^2{x}} \\ &= \frac{1}{2}\left(\int\frac{d(\cos{x})}{\cos{x} - 1} + \int\frac{d(\cos{x})}{\cos{x} + 1}\right) \\ &= \frac{1}{2}\ln\left\lvert\frac{\cos{x} - 1}{\cos{x} +1}\right\rvert + C \end{aligned}\]由于不定积分 \(\int\csc^2{x}dx = \cot{x} + C\)，\(\int\csc{t}\cot{x}dx = -\csc{x} + C\) ，原式可以转化为：
\[\begin{aligned} \int\csc{x}dx &= \int\frac{\csc{x}(\csc{x} - \cot{x})}{\csc{x} - \cot{x}}dx \\ &= \int\frac{d(\csc{x} - \cot{x})}{\csc{x} - \cot{x}} \\ &= \ln\lvert\csc{x} - \cot{x}\rvert + C \end{aligned}\]而因为 \(\csc{x} - \cot{x} = \frac{1 - \cos{x}}{\sin{x}} = \tan{\frac{x}{2}}\)，还可以推出：
\[\int\csc{x}dx = \ln\left\lvert\tan{\frac{x}{2}}\right\rvert + C \]（3）\(\int\frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan{\frac{x}{a}} + C\)
考虑[[换元积分法]]，设 \(x = a\tan{t}\)，故有：
\[\begin{aligned} \int\frac{dx}{x^2 + a^2} &= \int\frac{d(a\tan{t})}{a^2\tan^2{t} + a^2} \\ &= \frac{1}{a}\int\frac{d(\tan{t})}{\tan^2{t} + 1} \\ &= \frac{1}{a}\arctan(\tan{t}) + C \\ &= \frac{1}{a}\arctan\frac{x}{a} + C \end{aligned}\]（4）\(\int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left\lvert\frac{x - a}{x + a}\right\rvert + C\)
利用平方差公式，可以直接拆项：
\[\begin{aligned} \int\frac{dx}{x^2 - a^2} &= \frac{1}{2a}\left(\int\frac{dx}{x - a} - \int\frac{dx}{x + a}\right) \\ &= \frac{1}{2a}\left(\ln\lvert x - a\rvert - \ln\lvert x + a\rvert\right) + C \\ &= \frac{1}{2a}\left\lvert\frac{x - a}{x + a}\right\rvert + C \end{aligned}\]（5）\(\int\frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C\)
考虑换元积分法，设 \(x = a\sin{t}\)，故有：
\[\begin{aligned} \int\frac{dx}{\sqrt{a^2 - x^2}} &= \int\frac{d(a\sin{t})}{\sqrt{a^2 - a^2\sin^2{t}}} \\ &= \int\frac{d(\sin{t})}{\sqrt{1 - \sin^2{t}}} \\ &= \arcsin(\sin{t}) + C \\ &= \arcsin{\frac{x}{a}} + C \end{aligned}\]（6）\(\int\frac{dx}{\sqrt{x^2 + a^2}} = \ln\left(x + \sqrt{x^2 + a^2}\right) + C\)
考虑换元积分法，设 \(x = a\tan{t}\)，故有：
\[\begin{aligned} \int\frac{dx}{\sqrt{x^2 + a^2}} &= \int\frac{d(a\tan{t})}{\sqrt{a^2\tan^2{t} + a^2}} \\ &= \int\frac{d(\tan{t})}{\sqrt{\sec^2{t}}} \\ &= \int\lvert\sec{t}\rvert dt \end{aligned}\]根据公式（1）可以得到：
\[\int\lvert\sec{t}\rvert dt = \ln\lvert\sec{t} + \tan{t}\rvert + C \]不妨考虑 \(\tan{t} = \frac{x}{a}\) 的一个直角三角形，容易得出 \(\sec{t} = \frac{\sqrt{x^2 + a^2}}{a}\)，带入原式有：
\[\int\frac{dx}{\sqrt{x^2 + a^2}} = \ln\left\lvert\frac{\sqrt{x^2 + a^2}}{a} + \frac{x}{a}\right\rvert + C \]因为 \(\ln{a}\) 是一个常数，不妨取出后与 \(C\) 合并，即：
\[\int\frac{dx}{\sqrt{x^2 + a^2}} = \ln\left\lvert x + \sqrt{x^2 + a^2}\right\rvert + C \]（7）\(\int\frac{dx}{\sqrt{x^2 - a^2}} = \ln\left\lvert x + \sqrt{x^2 - a^2}\right\rvert + C\)
考虑换元积分法，设 \(x = a\sec{t}\)，故有：
\[\begin{aligned} \int\frac{dx}{\sqrt{x^2 - a^2}} &= \int\frac{d(a\sec{t})}{\sqrt{a^2\sec{t} - a^2}} \\ &= \int\frac{d(\sec{t})}{\sqrt{\tan^2{t}}} \\ &= \int\frac{\sec{t}\tan{t}}{\lvert\tan{t}\rvert}dt \\ &= \int\sec{t}dt \\ &= \ln\lvert\sec{t} + \tan{t}\rvert + C \end{aligned}\]不妨考虑 \(\sec{t} = \frac{x}{a}\) 的一个直角三角形，容易得出 \(\tan{t} = \frac{\sqrt{x^2 - a^2}}{a}\)，带入原式有：
\[\int\frac{dx}{\sqrt{x^2 - a^2}} = \ln\left\lvert\frac{x}{a} + \frac{\sqrt{x^2 - a^2}}{a}\right\rvert + C \]因为 \(\ln{a}\) 是一个常数，不妨取出后与 \(C\) 合并，即：
\[\int\frac{dx}{\sqrt{x^2 - a^2}} = \ln\left\lvert x + \sqrt{x^2 - a^2}\right\rvert + C \]积分表（对照查询）
为对齐美观，将常函数的积分规则放至最后。
| $$\int x^adx = \frac{x^{a + 1}}{a + 1} + C , (a \neq -1)$$ | $$\int \frac{dx}{x} = \ln{\lvert x \rvert} + C$$ |
|---|---|
| $$\int a^xdx = \frac{a^x}{\ln{a}} + C$$ | $$\int e^xdx = e^x + C$$ |
| $$\int\cos{x}dx = \sin{x} + C$$ | $$\int\sin{x}dx = -\cos{x} + C$$ |
| $$\int\sec^2{x}dx = \tan{x} + C$$ | $$\int\csc^2{x}dx = -\cot{x} + C$$ |
| $$\int\sec{x}\tan{x}dx = \sec{x} + C$$ | $$\int\csc{x}\cot{x}dx = -\csc{x} + C$$ |
| $$\int\frac{dx}{\sqrt{1 - x^2}} = \arcsin{x} + C$$ | $$\int-\frac{dx}{\sqrt{1 - x^2}} = \arccos{x} + C$$ |
| $$\int\frac{dx}{1 + x^2} = \arctan{x} + C$$ | $$\int-\frac{dx}{1 + x^2} = \operatorname{arccot}x + C$$ |
| $$\int\sinh{x} = \cosh{x} + C$$ | $$\int\cosh{x} = \sinh{x} + C$$ |
| $$\int\tan{x}dx = -\ln\lvert\cos{x}\rvert + C$$ | $$\int\cot{x}dx = \ln\lvert\sin{x}\rvert + C$$ |
| $$\int\sec{x}dx = \ln\lvert\sec{x} + \tan{x}\rvert + C$$ | $$\int\csc{x}dx = \ln\lvert\csc{x} - \cot{x}\rvert + C = \ln\left\lvert\tan{\frac{x}{2}}\right\rvert + C$$ |
| $$\int\frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan{\frac{x}{a}} + C$$ | $$\int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left\lvert\frac{x - a}{x + a}\right\rvert + C$$ |
| $$\int\frac{dx}{\sqrt{x^2 - a^2}} = \ln\left\lvert x + \sqrt{x^2 - a^2}\right\rvert + C$$ | $$\int\frac{dx}{\sqrt{x^2 + a^2}} = \ln\left(x + \sqrt{x^2 + a^2}\right) + C$$ |
| $$\int\frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C$$ | $$\int kdx = kx + C , (k \in C)$$ |
标签：frac,rvert,int,不定积分,常见,lvert,ln,dx From： https://www.cnblogs.com/YipChipqwq/p/18624663