Question

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

Answer 1

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

$=\int \frac{x\sqrt{1+\frac{1}{x^2}}[\log (x^2+1)-2\log x]}{x^4}dx$ [ traking $x^2$ common from $\sqrt{x^2+1}$ ]

$=\int \frac{{(1+\frac{1}{x^2})}^{1/2}\log (1+\frac{1}{x^2})}{x^3}dx$

Let $t=1+\frac{1}{x^2}$

$\Rightarrow \frac{dt}{dx}=\frac{-2}{x^3}$

$\Rightarrow \frac{dx}{x^3}=\frac{-dt}{2}$

Substituting

$=\frac{-1}{2}\int t^{1/2}\log tdt$

$=-\frac{1}{2}(\log t\int t^{1/2}dt-\int \frac{d}{dt}\left(\log t\right).(\int t^{\frac{1}{2}}dt)t)$

$=-\frac{1}{2}(\log t.\frac{2t^{\frac{3}{2}}}{3}-\frac{2}{3}\int t^{\frac{1}{2}}dt)$

$=-\frac{1}{2}(\log t.\frac{2t^{\frac{3}{2}}}{3}-\frac{2}{3}.\frac{2}{3}{t}^{\frac{3}{2}})$

$=\frac{2}{9}{t}^{\frac{3}{2}}-\frac{1}{3}{t}^{\frac{3}{2}}\log t$

Putting value of $t$

$=\frac{2}{9}{(1+\frac{1}{x^2})}^{\frac{3}{2}}-\frac{1}{3}{(1+\frac{1}{x^2})}^{\frac{3}{2}}\log (1+\frac{1}{x^2})+c$

$\therefore$ $-\frac{1}{3}{(1+\frac{1}{x^2})}^{\frac{3}{2}}\left(\log (1+\frac{1}{x^2})-\frac{2}{3}\right)+c$

Hence, the answer is $-\frac{1}{3}{(1+\frac{1}{x^2})}^{\frac{3}{2}}\left(\log (1+\frac{1}{x^2})-\frac{2}{3}\right)+c.$