Question

Integrate the function $\frac{\sqrt{x^2+1}\left[\log \right(x^2+1)-2\log x]}{x^4}$

Answer 1

$\frac{\sqrt{x^2+1}\left[\log \right(x^2+1)-2\log x]}{x^4}=\frac{\sqrt{x^2+1}}{x^4}\left[\log \right(x^2+1)-\log x^2]$
$=\frac{\sqrt{x^2+1}}{x^4}\left[\log \left(\frac{\sqrt{x^2+1}}{x^2}\right)\right]$
$=\frac{\sqrt{x^2+1}}{x^4}\log (1+\frac{1}{x^2})$
$=\frac{1}{x^3}\sqrt{\frac{\sqrt{x^2+1}}{x^2}}\log (1+\frac{1}{x^2})$
$=\frac{1}{x^3}\sqrt{1+\frac{1}{x^2}}\log (1+\frac{1}{x^2})$
Let $1+\frac{1}{x^2}=t\Rightarrow \frac{-2}{x^3}dx=dt$
$\therefore I=\int \frac{1}{x^3}\sqrt{1+\frac{1}{x^2}}\log (1+\frac{1}{x^2})dx$
$=-\frac{1}{2}\int \sqrt{t}\log tdt$
$=-\frac{1}{2}\int t^{\frac{1}{2}}\cdot \log tdt$
Integrating by parts, we obtain
$I=-\frac{1}{2}[\log t\cdot \int t^{\frac{1}{2}}dt-\{\left(\frac{d}{dt}\log t\right)\int t^{\frac{1}{2}}dt\}dt]$
$=-\frac{1}{2}[\log t\cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}-\int \frac{1}{t}\cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}dt]$
$=-\frac{1}{2}[\frac{2}{3}{t}^{\frac{3}{2}}\log t-\frac{2}{3}\int t^{\frac{1}{2}}dt]$
$=-\frac{1}{2}[\frac{2}{3}{t}^{\frac{3}{2}}\log t-\frac{4}{9}{t}^{\frac{3}{2}}]$
$=-\frac{1}{3}{t}^{\frac{3}{2}}\log t+\frac{2}{9}{t}^{\frac{3}{2}}$
$=-\frac{1}{3}{t}^{\frac{3}{2}}[\log t-\frac{2}{3}]$
$=-\frac{1}{3}{(1+\frac{1}{x^2})}^{\frac{3}{2}}[\log (1+\frac{1}{x^2})-\frac{2}{3}]+C$