Question

If $\int \frac{x^2}{\sqrt{1-4x^6}}dx=\frac{1}{6}{\sin }^{-1}\left\{f\right(x\left)\right\}+C$ then, $f^{\prime \prime }\left(1\right)$ is
(where $C$ is constant of integration)

Answer 1

put $x^3=y\Rightarrow 3x^{2}dx=dy$
$\therefore I=\frac{1}{3}\cdot \int \frac{dy}{\sqrt{1-4y^2}}=\frac{1}{3}\cdot \frac{{\sin }^{-1}\left(2y\right)}{2}+C$
$=\frac{1}{6}{\sin }^{-1}\left(2x^3\right)+C\Rightarrow f\left(x\right)=2x^3$
$f^{\prime }\left(x\right)=6x^2\Rightarrow f^{\prime \prime }\left(x\right)=12x\Rightarrow f^{\prime \prime }\left(1\right)=12$