<s>[INST] Integrate $\int \frac{x^3 e^x \:dx}{\sqrt{x^2-1}}$ Integrate $$\int \frac{x^3 e^x \:dx}{\sqrt{x^2-1}}$$
My try:
I used Parts by writing integral as:
$$I=\int x^2 e^x {\frac{x\:dx}{\sqrt{x^2-1}}}$$ we get
$$I=x^2 e^x\sqrt{x^2-1}-\int (x^2-2x+2)e^x \sqrt{x^2-1}dx$$
But the other integral has become more complicated
 [/INST] This is a modification which I hope will be helpful.
$$ \int (x^2 -2x+2)e^x \sqrt{ x^2-1}$$
$$\cot(cosec^{-1} x)=\sqrt{ x^2-1}$$
Let $cosec^{-1} x=u$ then:
$ cosec (u)=\frac{1}{\sin u} =x$
$dx=\frac{-\cos u}{\sin^2 u}du$
substituting we get:
$$ \int (x^2 -2x+2)e^x \sqrt{ x^2-1}=\int(\frac{1}{\sin^2 u}-\frac{2}{\sin u} +2)cotan (u) .\frac{-\cos u}{\sin^2 u} e^{1/\sin u}du$$
$\frac{-\cos u}{\sin^2 u} e^{1/\sin u}=(e^{1/\sin x})'$
$\frac{-\cos u}{\sin^2 u} e^{1/\sin u}du= dv$
$v=\frac{1}{\sin u}$
$(\frac{1}{\sin^2 u}-\frac{2}{\sin u} +2)cotan (u)=t$
I think you could now do it .
</s>