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 :)
 A: As @CameronWilliams suggested,
$$ x^2 -3 =t $$
and then
$$ x = \sqrt{t+3} $$
(I think you tried to find this in order to get the relation between dt and dx)
A: Another suggestion: for the case $x>0$, the hyperbolic substitution
$$x=\sqrt 3\cosh t,\qquad \mathrm dx=\sqrt 3\sinh t\,\mathrm dt$$
and similar for the case $x<0$.
With some hyperbolic trigonometry, you'll obtain a monomial in $\cosh t$ and $\sinh t$ for the integrand, which you'll have to linearise.
A: The domain of the integrand function is
$$(-\infty,-\sqrt{3}]\cup [\sqrt{3},+\infty)=I\cup J$$
to get the antiderivative at $ I $, put
$$x=\sqrt{3}\cosh(t).$$
and, at $ J $, make the substitution
$$x=-\sqrt{3}\cosh(t)$$
A: 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
$$I_0= \int \sqrt{x^2-3}\ dx=\frac x2\sqrt{x^2-3}-\frac32\tanh^{-1}\frac{\sqrt{x^2-3}}x
$$
