Question

Evaluate
$\int \frac{\sqrt{1+x^2}}{x}dx$

Answer 1

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