Integrate $\int x^4\sqrt{x^2-3} \, dx$

Question

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 :)

Answer 1

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)

Answer 2

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.

Answer 3

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 $$x=-\sqrt{3}\cosh(t)$$

Answer 4

Apply the reduction formula

$$\int x^n\sqrt{x^2-3}\ dx=I_n=\frac{x^{n-1}}{n+2}(x^2-3)^{3/2}+\frac{3(n-1)}{n+2}I_{n-2} $$ to reduce the integral to $I_0$

\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 $.