小学数学（Latex练习器）
\[\int \frac{\mathrm{d} x}{ax+b}=\frac{1}{a}\int \frac{\mathrm{d} (ax+b)}{ax+b}=\frac{1}{a}\ln|ax+b|+C
\]
\[\int (ax+b)^\mu\mathrm{d} x=\frac{1}{a}\int (ax+b)^\mu\mathrm{d} (ax+b)=\frac{1}{a(\mu+1)}(ax+b)^{\mu+1}+C(\mu\neq -1)
\]
\[\int \frac{x}{ax+b}\mathrm{d} x=\int \frac{1}{a}-\frac{b}{a(ax+b)}\mathrm{d} x=\frac{x}{a}-\frac{b}{a^2}\ln|ax+b|+C
\]
\[\int \frac{1}{x(ax+b)}\mathrm{d} x=\frac{1}{a}\int \frac{1}{x(x+\frac{b}{a})}\mathrm{d} x
=\frac{1}{b}\int \frac{1}{x}-\frac{1}{(x+\frac{b}{a})}\mathrm{d} x
=\frac{1}{b}(\ln |x|-\ln |x+\frac{b}{a}|+C_1)
=\frac{1}{b}(\ln |x|-\ln |ax+b|+C)
=\frac{1}{b}\ln\left|\frac{x}{ax+b}\right|+C
\]
\[\int \frac{1}{x^2+a^2}\mathrm{d} x=\frac{1}{a}\int \frac{1}{\frac{x^2}{a^2}+1}\mathrm{d} \frac{x}{a}=\frac{1}{a}\arctan \frac{x}{a}+C
\]
\[\int \tan x\mathrm{d} x=-\int \frac{1}{\cos x}\mathrm{d}\cos x=-\ln|\cos x|+C
\]
\[\int \arctan x\mathrm{d} x=x\arctan x-\int \frac{x}{1+x^2}\mathrm{d} x=x\arctan x-\frac{1}{2}\int \frac{1}{1+x^2}\mathrm{d} (1+x^2)
=x\arctan x-\frac{1}{2}\ln(1+x^2)+C;
\]
咕咕咕。。。
