一类 $\dif (ax+\frac{b}{x})$ 的不定积分
例题1：求不定积分: \(\displaystyle\int\frac{1}{1+x^4}\,\mathrm{d}x\)
解：
\[\begin{align*}
\int\frac{1}{1+x^4}\mathrm{d}x
&=\frac{1}{2}\int\frac{x^2+1}{1+x^4}\mathrm{d}x-\frac{1}{2}\int\frac{x^2-1}{1+x^4}\mathrm{d}x\\
&=\frac{1}{2}\int\frac{1}{\left(x-\frac{1}{x}\right)^2+2}\mathrm{d}\left(x-\frac{1}{x}\right)-\frac{1}{2}\int\frac{1}{\left(x+\frac{1}{x}\right)^2-2}\mathrm{d}\left(x+\frac{1}{x}\right)\\
&=\frac{1}{2\sqrt{2}}\arctan\frac{x^2-1}{\sqrt{2}}+\frac{1}{4\sqrt{2}}\ln\left|\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right|+C
\end{align*}\]
例题2：求不定积分: \(\displaystyle\int\sqrt{\tan x}\dif x\)
解：
\[\begin{align*}
\int\sqrt{\tan x}\dif x
&\xlongequal{\sqrt{\tan x}=t}2\int\frac{t^2}{1+t^4}\dif t
=\int\frac{1+t^2}{1+t^4}\dif t-\int\frac{1-t^2}{1+t^4}\dif t\\
&=\int\frac{1}{(t-\frac{1}{t})^2+2}\dif\left(t- \frac{1}{t}\right)-\int\frac{1}{(t+\frac{1}{t})^2- 2}\dif\left(t+\frac{1}{t}\right)\\
&=\frac{\sqrt 2}{2}\arctan\left(\frac{t^2-1}{\sqrt 2t}\right)+ \frac{\sqrt 2}{4}\ln\left|\frac{t^2+\sqrt 2t+1}{t^2-\sqrt 2t+1}\right| + c\\
&=\frac{\sqrt 2 }{2}\arctan\left(\frac{\tan x-1}{2\sqrt{\tan x}} \right) +\frac{\sqrt 2}{4}\ln\left|\frac{\tan x+\sqrt{2\tan x}+1}{\tan x-\sqrt{2\tan x}+1}\right| + c
\end{align*}\]
例题3：求不定积分 \(\displaystyle\int\frac{1}{x^8+x^4+1}\dif x\)
解：
\[\begin{align*}
\int\frac{1}{x^8+x^4+1}\dif x&=\int\frac{1}{(x^8+2x^4+1)-x^4}\dif x\\
&=\int\frac{1}{(x^4+1)^2-x^4}\dif x\\
&=\int\frac{1}{\big[(x^4+1)-x^2\big]\big[(x^4+1)+x^4\big]}\dif x\\
&=\int\frac{1}{(x^2-x+1)(x^2+x+1)(x^4-x^2+1)}\dif x\\
&=\frac{1}{4}\int\frac{\dif x}{x^2-x+1}+\frac{1}{4}\int\frac{\dif x}{x^2+x+1}+\frac{1}{2}\int\frac{1-x^2}{x^4-x^2+1}\dif x\\
&=\frac{1}{4}\int\frac{\dif x}{\big(x-\frac{1}{2}\big)^2+\frac{3}{4}}+\frac{1}{4}\int\frac{\dif x}{\big(x+\frac{1}{2}\big)^2+\frac{3}{4}}+\frac{1}{2}\int\frac{\frac{1}{x^2}-1}{x^2+\frac{1}{x^2}-1}\dif x\\
&=\frac{1}{2\sqrt{3}}\arctan\frac{2x-1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}\arctan\frac{2x+1}{\sqrt{3}}-\frac{1}{2}\int\frac{\big(x+\frac{1}{x}\big)^2}{\big(x+\frac{1}{x}\big)^2-2}\\
&=\frac{\arctan\frac{2x-1}{\sqrt{3}}+\arctan\frac{2x+1}{\sqrt{3}}}{2\sqrt{3}}-\frac{1}{4\sqrt{2}}\ln\left(\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right)+C
\end{align*}\]
例题4：求不定积分：\(\displaystyle\int\ln\Big(x+\frac{1}{x}\Big)\dif x\).
解：法I. 一方面，
\[\begin{align*}
\int\Big(1-\frac{1}{x^2}\Big)\ln\Big(x+\frac{1}{x}\Big)\dif x
&=\int\ln\Big(x+\frac{1}{x}\Big)\dif\Big(x+\frac{1}{x}\Big)\\
&\xlongequal{\text{分部积分}}\Big(x+\frac{1}{x}\Big)\ln\Big(x+\frac{1}{x}\Big)-\Big(x+\frac{1}{x}\Big)+C
\end{align*}\]
另一方面，
\[\begin{align*}
\int\Big(1+\frac{1}{x^2}\Big)\ln\Big(x+\frac{1}{x}\Big)\dif x
&=\frac{1}{2}\int\Big(1+\frac{1}{x^2}\Big)\ln\Big(x+\frac{1}{x}\Big)^2\dif x\\
&=\frac{1}{2}\int\Big(1+\frac{1}{x^2}\Big)\ln\bigg(\Big(x-\frac{1}{x}\Big)^2+4\bigg)\dif x\\
&=\frac{1}{2}\int\ln\bigg(\Big(x-\frac{1}{x}\Big)^2+4\bigg)\dif\Big(x-\frac{1}{x}\Big)\\
&\xlongequal{x-\frac{1}{x}=u}\frac{1}{2}\int\ln(u^2+4)\dif u\\
&\xlongequal{\text{分部积分}}\frac{1}{2}u\ln(u^2+4)+2\arctan\frac{u}{2}-u+C\\
&\xlongequal{\text{带值}}\Big(x-\frac{1}{x}\Big)\ln\Big(x+\frac{1}{x}\Big)+2\arctan\frac{x^2-1}{2x}-\Big(x-\frac{1}{x}\Big)+C
\end{align*}\]
所以
\[\int\ln\Big(x+\frac{1}{x}\Big)\dif x=x\ln\Big(x+\frac{1}{x}\Big)+\arctan\frac{x^2-1}{2x}-x+C
\]
法II. 直接分部积分
\[\begin{align*}
\int\ln\left(x+\frac{1}{x}\right){\rm d}x
&=x\ln\left(x+\frac{1}{x}\right)-\int\left(1-\frac{2}{x^2+1}\right){\rm d}x\\
&=x\ln\left(x+\frac{1}{x}\right)-x+2\arctan x+C.
\end{align*}\]
例题5：求不定积分：\(\displaystyle\int\frac{\dif x}{\sqrt{(x+x^{-1})^2-12}}\).
解：一方面，
\[\begin{align*}
\int\frac{1-\frac{1}{x^2}}{\sqrt{(x+x^{-1})^2-12}}\dif x
&=\int\frac{1}{\sqrt{(x+x^{-1})^2-12}}\dif\Big(x+\frac{1}{x}\Big)\\
&\xlongequal{x+\frac{1}{x}=u}\int\frac{1}{\sqrt{u^2-12}}\dif u\\
&=\ln|u+\sqrt{u^2-12}|+C%\\&\xlongequal{\text{带值}}\cdots
\end{align*}\]
另一方面，
\[\begin{align*}
\int\frac{1+\frac{1}{x^2}}{\sqrt{(x+x^{-1})^2-12}}\dif x
&=\int\frac{1}{\sqrt{(x-x^{-1})^2-8}}\dif\Big(x-\frac{1}{x}\Big)\\
&\xlongequal{x-\frac{1}{x}=v}\int\frac{1}{\sqrt{v^2-8}}\dif v\\
&=\ln|v+\sqrt{v^2-10}|+C%\\&\xlongequal{\text{带值}}\cdots
\end{align*}\]
所以
\[\int\frac{\dif x}{\sqrt{(x+x^{-1})^2-12}}=\frac{1}{2}\big(\ln|u+\sqrt{u^2-12}|+\ln|v+\sqrt{v^2-8}|\big)+C
\]
其中，\(u=x+\frac{1}{x},v=x-\frac{1}{x}\)
