Evaluating the indefinite integral $\int \frac{x^3}{\sqrt{4x^2 -1}}~dx$. 
Find $$\int {x^3\over \sqrt{4x^2 -1}}\,dx.$$


Let
$2x = \sec u$, $2 =\sec (u) \tan(u) u^{'}(x).$  Then
$$
\begin{align*}
\int {x^3\over \sqrt{4x^2 -1}}\,dx &= \frac1{16}\int {\sec^3 u\over \tan u}\tan u \sec u \,  du\\
 &= \frac1{16}\int {\sec^4 u} \, du\\ 
&= \frac1{16}\left(\tan u + {\tan^3 u\over 3} \right) + C\\
 &= \frac1{16}\left(\tan (\sec^{-1} 2x) + {\tan^3 (\sec^{-1} 2x)\over 3} \right) + C. 
\end{align*}$$
Given answer : $$\dfrac{\left(2x^2+1\right)\sqrt{4x^2-1}}{24}+C$$
Why is my answer incorrect ?

There is nothing wrong with your approach; you just need to fiddle around with trig identities to see that $\tan(\sec^{-1}2x)=\sqrt{\sec^2(\sec^{-1}2x)-1}=\sqrt{(2x)^2-1}$, etc.  But another, possibly easier, way to do the integral is to let $u=4x^2$ so that $du=8x\,dx$ and thus
$$\int{x^3\over\sqrt{4x^2-1}}dx={1\over32}\int{u\over\sqrt{u-1}}du={1\over32}\int\left(\sqrt{u-1}+{1\over\sqrt{u-1}}\right)du\\
={1\over48}(u-1)^{3/2}+{1\over16}(u-1)^{1/2}+C=\sqrt{u-1}\left(u+2\over48\right)+C\\
=\sqrt{4x^2-1}\left(4x^2+2\over48\right)+C=\sqrt{4x^2-1}\left(2x^2+1\over24\right)+C$$
