数学核心积分（综合）

$$\begin{aligned}&= e \int e^{-(x+1)}(x+1) dx \ &= e\left(-e^{-(x+1)}(x+1) - e^{-(x+1)}\right)\end{aligned}$$

$$\begin{aligned}&= 6 \int x^{-2} \ln(x) dx \ &= 6\left(-x^{-1}\ln x - x^{-1}\right)\end{aligned}$$

$$\begin{aligned}&= \frac{x^2}{2} \cdot \arctan(x) - \int \frac{x^2}{2} \cdot \frac{1}{1+x^2} dx \ &= \frac{x^2}{2} \cdot \arctan(x) - \int \left(\frac{1}{2} + \frac{-1}{1+x^2}\right) dx \ &= \frac{x^2}{2} \cdot \arctan(x) - \frac{1}{2}x + \frac{1}{2}\arctan(x)\end{aligned}$$

设：
$ dv = e^{\sin x} \cos x \, dx $，$ u = \sin x $

$$\begin{aligned}&= \sin x \cdot e^{\sin x} - \int e^{\sin x} \cos x \, dx \ &= (\sin x - 1)e^{\sin x}\end{aligned}$$

设：
$ u = x $，$ dv = (1-x)^6 dx $

$$\begin{aligned}&= x \cdot \frac{(1-x)^7}{-7} - \int \frac{(1-x)^7}{-7} dx \ &= x \cdot \frac{(1-x)^7}{-7} - \frac{(1-x)^8}{56}\end{aligned}$$

$$\begin{aligned}&= \ln(e^x + 1)\end{aligned}$$

$$\begin{aligned}&= \int \frac{1}{2\sin^2(x)} dx \ &= -\frac{1}{2}\cot(x) + C\end{aligned}$$

$$\begin{aligned}&= \int (\sec x)^2 (1 + (\tan x)^2) dx \ &= \int (\sec x)^2 dx + \int (\sec x)^2 (\tan x)^2 dx \ &= \tan x + \frac{(\tan x)^3}{3} + C\end{aligned}$$

$$\begin{aligned}&= \int \left(3 + \frac{2x + 3}{x^2 + 2x + 2}\right) dx \ &= \int \left(3 + \frac{2x + 2}{x^2 + 2x + 2} + \frac{1}{x^2 + 2x + 2}\right) dx \ &= 3x + \ln(x^2 + 2x + 2) + \arctan(x + 1) + C\end{aligned}$$

$$\begin{aligned}&= \int \tan x \cdot \sec^3 x \, dx \ &= \int \sec^2 x \cdot \sec x \tan x \, dx \ &= \frac{\sec^3 x}{3} + C\end{aligned}$$

$$\begin{aligned}&= \int \left(\frac{1}{1 + x} - \frac{1}{(1 + x)^2}\right) dx \ &= \ln(1 + x) + \frac{1}{1 + x} + C\end{aligned}$$

$$\begin{aligned}&= \int 2\cos^2 x \sin x \, dx \ &= -\frac{2}{3} \cos^3 x + C\end{aligned}$$

$$\begin{aligned}&= \int (1 - \sin^2(x))\cos(x) \, dx \ &= \sin(x) - \frac{\sin^3(x)}{3}\end{aligned}$$

$$\begin{aligned}&= \int (1 - \cos^2(x))\sin(x) \, dx \ &= -\cos(x) + \frac{\cos^3(x)}{3}\end{aligned}$$

$$\begin{aligned}&= \int (\sec^2(x) - 1)\tan(x) \, dx \ &= \frac{\tan^2(x)}{2} + \ln|\cos(x)|\end{aligned}$$

$$\begin{aligned}&= \int \frac{\cos(2x) + 1}{2} \, dx \ &= \frac{\sin(2x)}{4} + \frac{x}{2}\end{aligned}$$

$$\begin{aligned}&= \int \frac{1 - \cos(2x)}{2} \, dx \ &= \frac{x}{2} - \frac{\sin(2x)}{4}\end{aligned}$$

$$\begin{aligned}&= \tan(x)\end{aligned}$$

$$\begin{aligned}&= \int \sec^2(x) - 1 \, dx \ &= \tan(x) - x\end{aligned}$$

$$\begin{aligned}&= \int (\sec^2(x) - 1)\tan(x) \, dx \ &= \frac{\tan^2(x)}{2} + \ln|\cos(x)|\end{aligned}$$

$$\begin{aligned}&= \int (\sec^2(x) - 1)\tan^2(x) \, dx \ &= \frac{\tan^3(x)}{3} + \tan(x) - x\end{aligned}$$

$$\begin{aligned}&= \int (1 - \sin^2(x))\cos^2(x) \, dx \ &= \int \cos^2(x) \, dx - \int \sin^2(x)\cos^2(x) \, dx \ &= \frac{\sin(2x)}{4} + \frac{x}{2} - \int \left(\frac{\sin(2x)}{2}\right)^2 \, dx \ &= \frac{\sin(2x)}{4} + \frac{x}{2} - \frac{1}{4} \int \sin^2(2x) \, dx \ &= \frac{\sin(2x)}{4} + \frac{x}{2} - \frac{1}{4} \int \frac{1 - \cos(4x)}{2} \, dx \ &= \frac{\sin(2x)}{4} + \frac{x}{2} - \frac{1}{8}x + \frac{\sin(4x)}{32}\end{aligned}$$

$$\begin{aligned}&= \displaystyle \\ &= \int\left(\frac{1}{\cos^2 x}+\frac{\sin x}{\cos^2 x}\right)\mathrm{d}x \\ &= \int\left(\sec^2 x+\sec x\tan x\right)\mathrm{d}x \\ &= \tan x+\sec x+C.\end{aligned}$$

注意： 逐项对应 $\dfrac{\mathrm{d}}{\mathrm{d}x}(\tan x)$ 与 $\dfrac{\mathrm{d}}{\mathrm{d}x}(\sec x)$。

$$\begin{aligned}\int \frac{\sin(\ln x)}{x}\,\mathrm{d}x &= \int \sin(\ln x)\,\mathrm{d}(\ln x) \ &= -\cos(\ln x) + C\end{aligned}$$ 注意：此题考察了通过识别函数及其导数来简化积分的技巧。

$$\begin{aligned}\int \sqrt{9-x^2}\,\mathrm{d}x &= \int 3\sqrt{1-\left(\frac{x}{3}\right)^2}\,\mathrm{d}x \ &= 9\int \sqrt{1-\sin^2\theta}\cos\theta\,\mathrm{d}\theta & (x = 3\sin\theta, \,\mathrm{d}x = 3\cos\theta\,\mathrm{d}\theta) \ &= 9\int \cos^2\theta\,\mathrm{d}\theta \ &= \frac{9}{2}\int (1+\cos 2\theta)\,\mathrm{d}\theta \ &= \frac{9}{2}\left(\theta + \frac{1}{2}\sin 2\theta\right) + C \ &= \frac{9}{2}\left(\arcsin\left(\frac{x}{3}\right) + \frac{x}{3}\sqrt{1-\left(\frac{x}{3}\right)^2}\right) + C\end{aligned}$$ 注意：本题展示了如何利用三角换元法处理根号下的二次式。

$$\begin{aligned}\int x\ln x\,\mathrm{d}x &= \frac{x^2}{2}\ln x - \int \frac{x^2}{2}\cdot\frac{1}{x}\,\mathrm{d}x \ &= \frac{x^2}{2}\ln x - \int \frac{x}{2}\,\mathrm{d}x \ &= \frac{x^2}{2}\ln x - \frac{x^2}{4} + C\end{aligned}$$ 注意：这里运用了分部积分法，选择合适的u和dv是关键。