Question

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

Answer 1

Answer: 12

$f(x)=\int \frac{1+x^4}{{\left(1-x^4\right)}^{3/2}}dx$
Take out $x^2$ from the $D^r$, it will come out as $x^3$
$\int \frac{x+\frac{1}{x^3}}{{\left(\frac{1}{x^2}-x^2\right)}^{3/2}}dx$
$\text{ Put }\frac{1}{x^2}-x^2=t^2-2\left(x+\frac{1}{x^3}\right)dx=2tdt$
$f(x)=-\int \frac{tdt}{t^3}dt=\frac{1}{t}+C=\frac{x}{\sqrt{1-x^4}}+C_1;\text{ as }f(0)=0\Rightarrow C_1=0$
$$\begin{gathered} \text{ Now }\int \frac{x}{\sqrt{1-x^4}}dx=\frac{1}{2}{\sin }^{-1}{x}^2+C_2 \\ <br/>\text{ but }g(0)=0\Rightarrow C_2=0 \end{gathered}$$
$g(x)=\frac{1}{2}{\sin }^{-1}{x}^2$
$\text{ Hence }g\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{12}\equiv \frac{\pi }{k}\Rightarrow k=12$
$\text{ Alternatively : }I=\int \frac{1-x^4+2x^4}{{\left(1-x^4\right)}^{3/2}}dx=\int \left(\frac{1}{{\left(1-x^4\right)}^{1/2}}+\frac{2x^4}{{\left(1-x^4\right)}^{3/2}}\right)dx$
$\text{ This is of the form }f(x)+x{f}^{\prime }(x)$
$\text{ hence }I=xf(x)=\frac{x}{\sqrt{1-x^4}}+C$