Question

Let $\int \frac{(1+x^4)dx}{(1-x^4{)}^{\frac{3}{2}}}=f\left(x\right)+c_1$ where f(0) = 0 and $\int f\left(x\right).dx=g\left(x\right)+c_2$ with g(0) = 0.If $g\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{2k}$, then value of k is___

Answer 1

$\int \frac{(1+x^4)dx}{(1-x^4{)}^{\frac{3}{2}}}=\int \frac{(x+\frac{1}{x^3})}{{(\frac{1}{x^2}-x^2)}^{\frac{3}{2}}}dx$
$Put\frac{1}{x^2}-x^2=t^2$
$\Rightarrow -2(x+\frac{1}{x^3})dx=2tdt$
$f\left(x\right)=-\int \frac{t}{t^3}dt=\frac{1}{t}+C$
$=\frac{x}{\sqrt{1-x^4}}+C_1$
$asf\left(0\right)=0\Rightarrow C_1=0$
Now, $\int \frac{x}{\sqrt{1-x^4}}dx=\frac{1}{2}si{n}^{-1}{x}^2+C_2$
$g\left(0\right)=0\Rightarrow C_2=0$
$g\left(x\right)=\frac{1}{2}si{n}^{-1}\left(x^2\right)$
$g\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{12}$