Question

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

Answer 1

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

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

Let $\log \left(\frac{x^2+1}{x^2}\right)=t.\Rightarrow 1+\frac{1}{x^2}=e^t$

$\frac{1}{1+\frac{1}{x^2}}(-\frac{2}{x^3})dx=dt\Rightarrow \frac{1}{e^t}(-\frac{2}{x^3})dx=dt$

$\frac{1}{e^t.x^3}dx=-\frac{1}{2}dt$

$\Rightarrow \int -\frac{1}{2}\sqrt{e^t}.e^t.tdt=-\frac{1}{2}\int t{e}^{\frac{3}{2}t}dt$

$\Rightarrow -\frac{1}{2}[t\frac{e^{\frac{3}{2}t}}{\frac{3}{2}}-\frac{4}{9}{e}^{\frac{3}{2}}t]=-\frac{t{e}^{\frac{3}{2}t}}{3}+\frac{2}{9}{e}^{\frac{3}{2}t}$

put value of t.