题目：计算积分
\(\begin{align} \\&(1)\int \frac{7 \sin x+\cos x}{3 \sin x+4 \cos x} \mathrm{~d} x \text {. } \\&(2) \int \frac{1+\sin x}{1+\sin x+\cos x} \mathrm{~d} x \end{align}\)
解答：
(1)
\[7 \sin x+\cos x=(3 \sin x+4 \cos x)-(3 \sin x+4 \cos x)^{\prime} \\ 原式 =\int \mathrm{d} x-\int \frac{\mathrm{d}(3 \sin x+4 \cos x)}{3 \sin x+4 \cos x} \ \ \ \\=x-\ln |3 \sin x+4 \cos x|+C\](2）
方法 1——仿照上例拆分
\(\begin{aligned} \text { 原式 } &=\int \frac{\frac{1}{2}(1+\sin x+\cos x)+\frac{1}{2}(\sin x-\cos x)+\frac{1}{2}}{1+\sin x+\cos x} \mathrm{~d} x \\ &=\frac{1}{2} \int \mathrm{d} x-\frac{1}{2} \int \frac{\cos x-\sin x}{1+\sin x+\cos x} \mathrm{~d} x+\frac{1}{2} \int \frac{1}{1+\sin x+\cos x} \mathrm{~d} x \\ &=\frac{1}{2} x-\frac{1}{2} \int \frac{\mathrm{d}(1+\sin x+\cos x)}{1+\sin x+\cos x}+\frac{1}{2} \int \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}+2 \cos ^{2} \frac{x}{2}} \mathrm{~d} x \\ &=\frac{1}{2} x-\frac{1}{2} \ln |1+\sin x+\cos x|+\frac{1}{2} \int \frac{1}{\tan \frac{x}{2}+1} \mathrm{~d} \tan \frac{x}{2} \\ &=\frac{1}{2} x-\frac{1}{2} \ln |1+\sin x+\cos x|+\frac{1}{2} \ln \left|\tan \frac{x}{2}+1\right|+C \end{aligned}\)
方法 2——万能代换
令 $u=\tan \frac{x}{2}$, 则 $x=2 \arctan u, \mathrm{~d} x=\frac{2}{1+u^{2}} \mathrm{~d} u, \sin x=\frac{2 u}{1+u^{2}}, \cos x=\frac{1-u^{2}}{1+u^{2}}$. \(\text { 原式 }=\int \frac{1+\frac{2 u}{1+u^{2}}}{1+\frac{2 u}{1+u^{2}}+\frac{1-u^{2}}{1+u^{2}}} \frac{2}{1+u^{2}} \mathrm{~d} u=\int \frac{u+1}{1+u^{2}} \mathrm{~d} u \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\=\arctan u+\frac{1}{2} \ln \left(1+u^{2}\right)+C=\frac{x}{2}+\ln \left|\sec \frac{x}{2}\right|+C . \\\)