Question

Let $\int \frac{(1+x^4)dx}{(1-x^4{)}^{3/2}}=f\left(x\right)+C_1$ with $f\left(0\right)=0$ and $\int f\left(x\right)dx=g\left(x\right)+C_2$ with $g\left(0\right)=0.$ If $g\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{k},$ then the value of $k$ is

Answer 1

$I_1=\int \frac{(1+x^4)}{(1-x^4{)}^{3/2}}dx$
$=\int \frac{x+\frac{1}{x^3}}{{(\frac{1}{x^2}-x^2)}^{3/2}}dx$
Put $\frac{1}{x^2}-x^2=t^2$
$\Rightarrow -2(x+\frac{1}{x^3})dx=2tdt$
Then, $I_1=-\int \frac{t}{t^3}dt$
$=\frac{1}{t}+C_1$
$=\frac{x}{\sqrt{1-x^4}}+C_1$
As $f\left(0\right)=0\Rightarrow C_1=0$

Now, $\int \frac{xdx}{\sqrt{1-x^4}}=\frac{1}{2}{\sin }^{-1}{x}^2+C_2$
But $g\left(0\right)=0\Rightarrow C_2=0$
$\therefore g\left(x\right)=\frac{1}{2}{\sin }^{-1}{x}^2$
Hence, $g\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{12}\equiv \frac{\pi }{k}$
$\Rightarrow k=12$