<s>[INST] Integrate $\int x^4\sqrt{x^2-3} \, dx$ Integrate $\int x^4\sqrt{x^2-3} \, dx$
I tried substituting $\sqrt{x^2-3}=t$ than by squaring both sides and by simplifying i got $x\,dx=t\,dt$ and $x^2=t^2+3$
Now after substituting to integral i have  $\int (t^2+3)^2\cdot t\frac{t\,dt}{x}$, can't get rid of $x$ :(
Don't know how to move on, need a bit help if possible.
Thanks you in advance :)
 [/INST] Utilize the recursive relation
$$\int x^n\sqrt{x^2-3}\ dx=I_n=\frac{x^{n-1}}{n+2}(x^2-3)^{3/2}+\frac{3(n-1)}{n+2}I_{n-2}
$$
to evaluate the integral systematically
\begin{align}
\int x^4\sqrt{x^2-3}\ dx=
\left(\frac16 x^3 +\frac38 x \right)(x^2-3)^{3/2}+\frac98I_0
\end{align}
where
$$I_0= \int \sqrt{x^2-3}\ dx=\frac x2\sqrt{x^2-3}-\frac32\tanh^{-1}\frac{\sqrt{x^2-3}}x
$$
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