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。
$$\int x^adx = \frac{x^{a + 1}}{a + 1} + C , (a \neq -1)$$ |
$$\int \frac{dx}{x} = \ln{\lvert x \rvert} + C$$ |
$$\int a^xdx = \frac{a^x}{\ln{a}} + C$$ |
$$\int e^xdx = e^x + C$$ |
$$\int\cos{x}dx = \sin{x} + C$$ |
$$\int\sin{x}dx = -\cos{x} + C$$ |
$$\int\sec^2{x}dx = \tan{x} + C$$ |
$$\int\csc^2{x}dx = -\cot{x} + C$$ |
$$\int\sec{x}\tan{x}dx = \sec{x} + C$$ |
$$\int\csc{x}\cot{x}dx = -\csc{x} + C$$ |
$$\int\frac{dx}{\sqrt{1 - x^2}} = \arcsin{x} + C$$ |
$$\int-\frac{dx}{\sqrt{1 - x^2}} = \arccos{x} + C$$ |
$$\int\frac{dx}{1 + x^2} = \arctan{x} + C$$ |
$$\int-\frac{dx}{1 + x^2} = \operatorname{arccot}x + C$$ |
$$\int\sinh{x} = \cosh{x} + C$$ |
$$\int\cosh{x} = \sinh{x} + C$$ |
$$\int\tan{x}dx = -\ln\lvert\cos{x}\rvert + C$$ |
$$\int\cot{x}dx = \ln\lvert\sin{x}\rvert + C$$ |
$$\int\sec{x}dx = \ln\lvert\sec{x} + \tan{x}\rvert + C$$ |
$$\int\csc{x}dx = \ln\lvert\csc{x} - \cot{x}\rvert + C = \ln\left\lvert\tan{\frac{x}{2}}\right\rvert + C$$ |
$$\int\frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan{\frac{x}{a}} + C$$ |
$$\int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left\lvert\frac{x - a}{x + a}\right\rvert + C$$ |
$$\int\frac{dx}{\sqrt{x^2 - a^2}} = \ln\left\lvert x + \sqrt{x^2 - a^2}\right\rvert + C$$ |
$$\int\frac{dx}{\sqrt{x^2 + a^2}} = \ln\left(x + \sqrt{x^2 + a^2}\right) + C$$ |
$$\int\frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C$$ |
$$\int kdx = kx + C , (k \in C)$$ |