Question

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

Answer 1

$LetI=\int \frac{\sqrt{x^2+1}\left[log\right(x^2+1)-2logx]}{x^4}dx=\int \frac{\sqrt{x^2+1}}{x^4}\left[log\right(x^2+1)-log{x}^2]dx=\int \frac{\sqrt{x^2+1}}{x^4}\left[log(1+\frac{1}{x^2})\right]dx=\int \frac{1}{x^3}\sqrt{\frac{x^2+1}{x^2}}\left[log(1+\frac{1}{x^2})\right]dxPut\frac{x^2+1}{x^2}=t\Rightarrow 1+\frac{1}{x^2}=t\Rightarrow -\frac{2}{x^3}dx=dt\Rightarrow dx=\frac{-x^3}{2}dt\therefore I=\int \sqrt{t}logt.\frac{1}{x^3}(-\frac{x^{3}dt}{2})=-\frac{1}{2}\int (t{)}^{1/2}logtdt$

On integrating by parts, we get

$I=-\frac{1}{2}[logt\int t^{\frac{1}{2}}dt-\int \{\left(\frac{d}{dt}logt\right)\int t^{\frac{1}{2}}dt\}dt]=-\frac{1}{2}[logt.\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-\int \frac{1}{t}.\frac{t^{\frac{3}{2}}}{\frac{3}{2}}dt]=-\frac{1}{2}[\frac{2}{3}logt.t^{\frac{3}{2}}-\frac{2}{3}\int t^{\frac{1}{2}}dt]=-\frac{1}{2}\times \frac{2}{3}[logt.t^{\frac{3}{2}}-\int t^{\frac{1}{2}}dt]=-\frac{1}{3}[logt.t^{\frac{3}{2}}-\frac{t^{\frac{3}{2}}}{\frac{3}{2}}]+C=-\frac{1}{3}{t}^{\frac{3}{2}}[logt-\frac{2}{3}]+C=-\frac{1}{3}{(1+\frac{1}{x^2})}^{\frac{3}{2}}[log\left(\frac{x^2+1}{x^2}\right)-\frac{2}{3}]+C$