If $\int \frac{\sqrt{1-x^4}}{x^7} d x=f(x)\left\{\sqrt{1-x^4}\right\}^n+C$, then $(f(x))^n$ is equal to
If $\int \frac{\sqrt{1-x^4}}{x^7} d x=f(x)\left\{\sqrt{1-x^4}\right\}^n+C$, then $(f(x))^n$ is equal to
$\frac{-1}{6 x^6}$
$\frac{-1}{216 x^{18}}$
$\frac{1}{36 x^{12}}$
$\frac{1}{216 x^{18}}$
Solution
$I=\int \frac{\sqrt{1-x^4}}{x^7} d x$
Let $x^2=u$, then $2 x d x=d u$
$
I=\int \frac{\sqrt{1-u^2}}{2 u^4} d u=\frac{1}{2} \int \frac{\sqrt{1-u^2}}{u^4} d u
$
Let $u=\sin v$, then $d u=\cos v d v$
$
\begin{aligned}
I & =\frac{1}{2} \int \frac{\cos ^2 v}{\sin ^4 v} d v=\frac{1}{2} \int \frac{1}{\cos ^2 v \tan ^4 v} d v \\
& =\frac{1}{2} \int \frac{\sec ^2 v}{\tan ^4 v} d v
\end{aligned}
$
Let $\tan v=w$, then $\sec ^2 v d v=d w$
$
\begin{aligned}
& \quad I=\frac{1}{2} \int \frac{d w}{w^4}=\frac{1}{2}\left(\frac{w^{-3}}{-3}\right)=\frac{-1}{6 w^3}+C \\
& =\frac{-1}{6 \tan ^3 v}=\frac{-\left(\sqrt{1-u^2}\right)^3}{6 u^3}=-\frac{1}{6} \frac{\left(1-x^4\right)^{3 / 2}}{x^6}+C \\
& \therefore \quad I=\frac{-1}{6} \frac{\left(1-x^4\right)^{3 / 2}}{x^6}+C
\end{aligned}
$
Now, $I=f(x)\left(1-x^4\right)^{n / 2}+C$
on Comparing, we obtain
$
\begin{gathered}
f(x)=\frac{-1}{6 x^6} \text { and } n=3 \\
\therefore \quad(f(x))^n=\left(-\frac{1}{6 x^6}\right)^3=\frac{-1}{216 x^{18}}
\end{gathered}
$