<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] $$\tan\left(\sec^{-1}(2x)\right)=\sqrt{4x^2-1}$$
Make this substitution into your answer and you will get the result you expected.
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