Question

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

Answer 1

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

$=\int \frac{\sqrt{x^2+1}\log \left(\frac{x^2+1}{x^2}\right)}{x^4}dx$

$=\int \sqrt{\frac{x^2+1}{x^2}}\log \left(\frac{x^2+1}{x^2}\right)\times \frac{1}{x^3}.dx$

put $\int \sqrt{\frac{x^2+1}{x^2}}=t$, $\frac{x^2+1}{x^2}=t^2$

$\frac{-2}{x^3}.dx=2tdt$

$+\frac{1}{x^3}dx=-tdt$

$=\int t.\log t^2(-t)dt$

$=-\int t^2\times \log \left(t^2\right)\times dt=[\log t^2\times \frac{t^3}{3}-\int \frac{1}{t^2}\times 2t\times \frac{t^3}{3}]$

$=-[\log t^2\times \frac{t^3}{3}-\frac{2}{3}\times \frac{t^3}{3}]+C$

$=-\frac{1}{3}{t}^3\log t^2+\frac{2}{3}{t}^3+C\Rightarrow \frac{-1}{3}{t}^3[\log t^2-\frac{2}{3}]+C$

$=\frac{-1}{3}{\left[\frac{x^2+1}{x^2}\right]}^{3/2}[\log \left(\frac{x^2+1}{x^2}\right)-\frac{2}{3}]+C$