Question

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

Answer 1

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

Apply the integration by parts,

$u=\sqrt{1+x^2},v=\frac{1}{x^2}$

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

$=-\frac{\sqrt{1+x^2}}{x}-(-\ln |\sqrt{x^2+1}+x|)$

$=-\frac{\sqrt{1+x^2}}{x}+\ln |\sqrt{x^2+1}+x|+C$

Hence, the required result is found.