Question

Solve $\int \left[\frac{\sqrt{x^2+1}\left[\ln (x^2+1)-2\ln x\right]}{x^4}\right]dx$

Answer 1

$\int \frac{\sqrt{x^2+1}\left[\ln \right(x^2+1)-2\ln x]}{x^4}dx=\int \sqrt{\frac{x^2+1}{x^2}}\times \ln \frac{x^2+1}{x^2}\times \frac{dx}{x^3}...\left(1\right)$

Put $t=\frac{x^2+1}{x^2}⟹dt=-\frac{2}{x^3}dx$

$\left(1\right)$ becomes $\int \sqrt{t}\ln t(-\frac{dt}{2})=-\frac{1}{2}[\ln t\int \sqrt{t}dt-\int \frac{d\left(\ln t\right)}{dt}\times \int \sqrt{t}dt\times dt]$

$=-\frac{1}{2}[\frac{2}{3}{t}^{\frac{3}{2}}\ln t-\int \frac{1}{t}\times \frac{2}{3}\times t^{\frac{3}{2}}dt]$

$=-\frac{1}{2}[\frac{2}{3}{t}^{\frac{3}{2}}\ln t-\frac{4}{9}{t}^{\frac{3}{2}}]=-\frac{1}{9}[3t^{\frac{3}{2}}\ln t-2t^{\frac{3}{2}}]$