不定积分专项练习

📌 重要
可能用到的一些超出高中数学的公式:
$\cot x = \cfrac{1}{\tan x}$
$\sec x = \cfrac{1}{\cos x}$
$\csc x = \cfrac{1}{\sin x}$
$\sec^2 x - \tan^2 x = 1$
$\csc^2 x - \cot^2 x = 1$
$\frac{\mathrm{d}}{\mathrm{d}x} \tan x = \sec^2 x$
$\frac{\mathrm{d}}{\mathrm{d}x} \cot x = - \csc^2 x$
$\frac{\mathrm{d}}{\mathrm{d}x} \sec x = \sec x \tan x$
$\frac{\mathrm{d}}{\mathrm{d}x} \csc x = - \csc x \cot x$

凑微分法

例题

求不定积分:
$$\int x \sin x^2 \mathrm{d}x $$

【解】: 把 $x$ 凑到 $\mathrm{d}$ 后面去

$$ \begin{aligned} I &= \cfrac{1}{2} \int \sin x^2 \mathrm{d}(x^2) \\ &= - \cfrac{1}{2} \cos x^2 + C \end{aligned} $$

练习

求解下列不定积分:

$ \displaystyle \int x e^{x^2} \mathrm{d}x $
$ \displaystyle \int \cfrac{\ln x}{x} \mathrm{d}x $
$ \displaystyle \int \sin^3 x \cos x \mathrm{d}x $
$ \displaystyle \int \cfrac{1}{\sqrt{1-x^2}} \cdot \cfrac{1}{\sqrt{1-(\arcsin x)^2}} \mathrm{d}x $
$ \displaystyle \int \cfrac{x}{(1+x^2)^2} \mathrm{d}x $
$ \displaystyle \int \cfrac{1}{x \ln x (\ln(\ln x))^2} \mathrm{d}x $
$ \displaystyle \int \cfrac{e^x (1 + \sin x)}{1 + \cos x} \mathrm{d}x $
$ \displaystyle \int \cfrac{1}{\sin^2 x \cos^2 x} \mathrm{d}x $
$ \displaystyle \int \cfrac{\sin x \cos x}{\sqrt{1 + \sin^4 x}} \mathrm{d}x $
$ \displaystyle \int \tan^4 x \mathrm{d}x $
$ \displaystyle \int \cfrac{x e^x}{(1+x)^2} \mathrm{d}x $
$ \displaystyle \int \cfrac{1 - \ln x}{(x - \ln x)^2} \mathrm{d}x $

代入换元法

例题

求不定积分:
$$\int \sqrt{1 - x^2} \mathrm{d}x $$

【解】: 看到 $\sqrt{1 - x^2}$, 考虑三角换元 $x = \sin t$, 则 $\mathrm{d}x = \cos t \mathrm{d}t$, 故

$$ \begin{aligned} I &= \int \sqrt{1 - \sin^2 t} \cos t \mathrm{d}t \\ &= \int \cos^2 t \mathrm{d}t \\ & = \cfrac{1}{2} \int (1 + \cos 2t) \mathrm{d}t \\ &= \cfrac{1}{2} t + \cfrac{1}{4} \sin 2t + C \end{aligned} $$

记得变量还原, 由 $x = \sin t$ 可得 $t = \arcsin x$, 又因为 $\sin 2t = 2 \sin t \cos t = 2x \sqrt{1 - x^2}$, 故

$$ I = \cfrac{1}{2} \arcsin x + \cfrac{1}{2} x \sqrt{1 - x^2} + C $$

练习

$ \displaystyle \int \sqrt{4 - x^2} dx $
$ \displaystyle \int \frac{1}{x^2 \sqrt{x^2 + 1}} dx $
$ \displaystyle \int \frac{1}{\sqrt{x} + \sqrt[3]{x}} dx $

综合练习

$ \displaystyle \int \frac{\mathrm{d}x}{1+e^{x}} $
$ \displaystyle \int \frac{\mathrm{d}x}{x\left(x^{6}+4\right)} $
$ \displaystyle \int \frac{\mathrm{d}x}{\sqrt{x(1-x)}} $
$ \displaystyle \int \frac{\mathrm{d}x}{x \sqrt{x^{2}-1}} $
$ \displaystyle \int \frac{\mathrm{d}x}{x \sqrt{4-x^{2}}} $
$ \displaystyle \int \sqrt{\frac{A+x}{A-x}} \mathrm{d}x $
$ \displaystyle \int \sqrt{\frac{x-A}{B-x}} \mathrm{d}x $
- 注: $A<B$
$ \displaystyle \int \frac{\mathrm{d}x}{1+x^{3}} $