Question

Solve: $\int \frac{x^2-1}{x\sqrt{1+x^4}}\text{dx}$

Answer 1

$\begin{array}{c}\int \frac{x^2-1}{x\sqrt{1+x^4}}dx\\ \Rightarrow \int \frac{x}{\sqrt{1+x^4}}-\int \frac{1}{x\sqrt{1+x^4}}dx\\ \Rightarrow \frac{1}{2}\int \frac{dt}{\sqrt{1+t^2}}-\frac{1}{2}\int \frac{1}{t\sqrt{1+t^2}}dt\\ \Rightarrow \frac{1}{2}\log \left|t+\sqrt{t^2+1}\right|\frac{-1}{2}\int t\frac{1}{\sqrt{\left(1+t^2\right)}}dt\\ let,-\sqrt{1+t^2}=P\\ \Rightarrow \frac{1}{2\sqrt{1+t^2}}.2tdt=dp\\ \Rightarrow \frac{1}{2}\log \left|x^2+\sqrt{x^4+1}\right|\frac{-1}{2}\int \frac{1}{\left(P^2-1\right)}dp\\ \Rightarrow \frac{1}{2}\log \left|x^2+\sqrt{1+x^4}-\frac{1}{2}\times \frac{1}{2}\log \right|\frac{P-1}{P+1}|+c\\ \Rightarrow \frac{1}{2}\log \left|x^2+\sqrt{1+x^4}\right|-\frac{1}{4}\log \left|\frac{P-1}{P+1}\right|+c\\ =\frac{1}{2}\log \left|x^2+\sqrt{x^4+1}-\frac{1}{4}\log \right|\frac{\sqrt{1+x^4}-1}{\sqrt{1+x^4}+1}|.\end{array}$