Integrate $\int\frac{1}{(x^2-3)\sqrt{x^2+2}}dx$

Question

Solve integral $\int\frac{1}{(x^2-3)\sqrt{x^2+2}}dx$

I tried substituting $x=tan(t)=\frac{sin(t)}{cos(t)}\:;\:dx=\frac{1}{cos^2(t)}dt$

So $\:{x^2+2}=\frac{sin^2(t)+2cos^2(t)}{cos^2(t)}=\frac{(1-cos^2(t))+2cos^2(t)}{cos^2(t)}=\frac{1+cos^2(t)}{cos^2(t)}=tan^2(t)$

$\sqrt{x^2+2}=tan(t)$

And $\:x^2-3=\frac{1-4cos^2(t)}{cos^2(t)}$

After putting both into main integral i have $\int\frac{cos(t)}{sin(t)\cdot(1-4cos^3(t))}dt=?$

Now need a bit help if possible.

p.s Is any other method or its fine with this sub. ?

Thank you in advance :)

Answer 1

If you do $\require{cancel}x=\sqrt2\tan(\theta)$ and $\mathrm dx=\sqrt2\sec^2(\theta)\,\mathrm d\theta$, then you will get$$\int\frac{\cancel{\sqrt2}\sec^2(\theta)}{\left(2\tan^2(\theta)-3\right)\cancel{\sqrt2}\sqrt{\tan^2(\theta)+1}}\,\mathrm d\theta.\tag1$$Since $1+\tan^2=\sec^2$, $(1)$ becomes\begin{align}\int\frac{\sec(\theta)}{\left(2\tan^2(\theta)-3\right)}\,\mathrm d\theta&=\int\frac{\cos(\theta)}{2\sin^2(\theta)-3\cos^2(\theta)}\,\mathrm d\theta\\&=\int\frac{\cos(\theta)}{5\sin^2(\theta)-3}\,\mathrm d\theta.\end{align}And now you can do $\sin(\theta)=y$ and $\cos(\theta)\,\mathrm d\theta=\mathrm dy$.

Answer 2

You have made a mistake. You write that $$x^2+2=\tan^2 t$$ but this is clearly false, since $x^2=\tan^2 t$! Your error is in the use of the identity $$1+\tan^2x\equiv \sec^2 x.$$ You mistakenly write that $$1+\sec^2 x\equiv\tan^2 x.$$ Once you have fixed this and had another bash at the integral I'll be happy to help if you're still stuck.