<s>[INST] 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 ?
 [/INST] 
By the above diagram, we can simplify the answer obtained by the author.
$$
\begin{array}{l}
\begin{aligned}
I &=\frac{1}{16}\left(\tan \theta+\frac{\tan ^{3} \theta}{3}\right)+C \\
&=\frac{1}{16}\left[\sqrt{4 x^{2}-1}+\frac{\left(4 x^{2}-1\right)^{\frac{3}{2}}}{3}\right]+C
\\&=\frac{\sqrt{4 x^{2}-1}}{48}\left[3+4 x^{2}-1\right]+C \\
\\&=\frac{\left(2 x^{2}+1\right) \sqrt{4 x^{2}-1}}{24}+C
\end{aligned} \\
\end{array}
$$
</s>