<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] Once you get to $\tfrac{1}{16}(\tan u + \tfrac{1}{3} \tan^3 u) + C$, get rid of $u$. Since $\sec u = 2x = 2x/1$, that means $\tan u = \sqrt{4x^2 - 1}$ (draw a right triangle to show this), so $$\begin{aligned}[t]\text{integral} = \tfrac{1}{16}(\tan u + \tfrac{1}{3} \tan^3 u) + C &= \tfrac{1}{16}\Big(\sqrt{4x^2 - 1} + \tfrac{1}{3} (4x^2-1)\sqrt{4x^2-1}\,\Bigr) + C \\ &= \tfrac{1}{16}\sqrt{4x^2-1} \, \Bigl(1 + \tfrac{4}{3} x^2 - \tfrac{1}{3} \Bigr) + C \\ &= \tfrac{1}{24}\sqrt{4x^2-1} \, \Bigl(2x^2+1 \Bigr) + C.\end{aligned}$$
</s>