# Show that

Question:

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

Solution:

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

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

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

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

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

Let $1+\frac{1}{x^{2}}=t \Rightarrow \frac{-2}{x^{3}} d x=d t$

\begin{aligned} \therefore I &=\int \frac{1}{x^{3}} \sqrt{1+\frac{1}{x^{2}}} \log \left(1+\frac{1}{x^{2}}\right) d x \\ &=-\frac{1}{2} \int \sqrt{t} \log t d t \\ &=-\frac{1}{2} \int t^{\frac{1}{2}} \cdot \log t d t \end{aligned}

Integrating by parts, we obtain

$I=-\frac{1}{2}\left[\log t \cdot \int t^{\frac{1}{2}} d t-\left\{\left(\frac{d}{d t} \log t\right) \int t^{\frac{1}{2}} d t\right\} d t\right]$

$=-\frac{1}{2}\left[\log t \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}-\int \frac{1}{t} \frac{t^{\frac{3}{2}}}{\frac{3}{2}} d t\right]$

$=-\frac{1}{2}\left[\frac{2}{3} t^{\frac{3}{2}} \log t-\frac{2}{3} \int t^{\frac{1}{2}} d t\right]$

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

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

$=-\frac{1}{3} t^{\frac{3}{2}}\left[\log t-\frac{2}{3}\right]$

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