Question

If $I=\displaystyle \int \frac{1 + x^{4}}{\left(1 - x^{4}\right)^{\frac{3}{2}}}dx$ $=\frac{1}{\sqrt{f \left(x\right)}}+C$ (where, $C$ is the constant of integration) and $f\left(2\right)=\frac{- 15}{4}$ , then the value of $2f\left(\frac{1}{\sqrt{2}}\right)$ is

Answer 1

Given integral is
$I=\displaystyle \int \frac{\frac{1}{x^{3}} + x}{\left(\frac{1}{x^{2}} - x^{2}\right)^{\frac{3}{2}}}dx$
Let $\frac{1}{x^{2}}-x^{2}=t$
$\Rightarrow \left(- \frac{2}{x^{3}} - 2 x\right)dx=dt$
$\therefore I=-\frac{1}{2}\displaystyle \int \frac{d t}{t^{\frac{3}{2}}}$
$=t^{- \frac{1}{2}}+C$
$=\frac{1}{\sqrt{\frac{1}{x^{2}} - x^{2}}}+C$
$\therefore f\left(x\right)=\frac{1}{x^{2}}-x^{2}$
$\Rightarrow f\left(\frac{1}{\sqrt{2}}\right)=2-\frac{1}{2}=\frac{3}{2}$
$\Rightarrow 2f\left(\frac{1}{\sqrt{2}}\right)=3$