Question

If $I=\int \frac{1+x^4}{{\left(1-x^4\right)}^{\frac{3}{2}}}dx$ $=\frac{1}{\sqrt{f\left(x\right)}}+C$ (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

Answer: 3

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