Question

Let $\int \frac{\left(1+x^4\right) d x}{\left(1-x^4\right)^{\frac{3}{2}}}=f(x)+C_1$ where $f(0)=0$ and $\int f(x) d x=g(x)+C_2$ with $g(0)=0$.
If $g \left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{ k }$. Find $k$.

Answer 1

$f(x)=\int \frac{1+x^4}{\left(1-x^4\right)^{3 / 2}} d x$
Take out $x ^2$ from the $D ^r$, it will come out as $x ^3$
$\int \frac{x+\frac{1}{x^3}}{\left(\frac{1}{x^2}-x^2\right)^{3 / 2}} d x $
$\text { Put } \frac{1}{x^2}-x^2=t^2-2\left(x+\frac{1}{x^3}\right) d x=2 t d t $
$f(x)=-\int \frac{t d t}{t^3} d t=\frac{1}{t}+C=\frac{x}{\sqrt{1-x^4}}+C_1 ; \text { as } f(0)=0 \Rightarrow C_1=0 $
$\text { Now } \int \frac{ x }{\sqrt{1- x ^4}} dx =\frac{1}{2} \sin ^{-1} x ^2+ C _2 \\
\text { but } g (0)=0 \Rightarrow C _2=0$
$g ( x )=\frac{1}{2} \sin ^{-1} x ^2$
$\text { Hence } g\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{12} \equiv \frac{\pi}{k} \Rightarrow k=12 $
$\text { Alternatively : } I=\int \frac{1-x^4+2 x^4}{\left(1-x^4\right)^{3 / 2}} d x=\int\left(\frac{1}{\left(1-x^4\right)^{1 / 2}}+\frac{2 x^4}{\left(1-x^4\right)^{3 / 2}}\right) d x$
$\text { This is of the form } f(x)+x f^{\prime}(x) $
$\text { hence } I=x f(x)=\frac{x}{\sqrt{1-x^4}}+C$