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

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

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

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

Now need a bit of help if possible.

p.s Is there any other method or is it 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.

Answer 3

Another method is to substitute $x=\frac1{t}$:

$$\mathcal I=\int\frac{\mathrm dx}{(x^2-3)\sqrt{x^2+2}}$$ $$=\text{sgn}t\int\frac{t\mathrm dt}{(3t^2-1)\sqrt{2t^2+1}}$$

Now, substitute $\sqrt{2t^2+1}=u$:

$$\mathcal I=\frac{\text{sgn}x}2\int\frac{\not{u}\mathrm du}{\not{u}\left(\frac32(u^2-1)-1\right)}$$ $$=\text{sgn}x\int\frac{\mathrm du}{3u^2-5}$$ $$=\frac{\text{sgn}x}{2\sqrt{15}}\ln\left|\frac{\sqrt5-\sqrt3 u}{\sqrt5+\sqrt3 u}\right|+C$$

$$\therefore\boxed{\int\frac{\mathrm dx}{(x^2-3)\sqrt{x^2+2}}=\frac{\text{sgn}x}{2\sqrt{15}}\ln\left|\frac{\sqrt5|x|-\sqrt3 \sqrt{2+x^2}}{\sqrt5|x|+\sqrt3 \sqrt{2+x^2}}\right|+C}$$

Answer 4

Setting $x=\dfrac1{\sqrt 2}\left(t-\dfrac1t\right)$ will rationalize the integrand, because $\sqrt{x^2+2}=\dfrac1{\sqrt 2}\left(t+\dfrac1t\right)$.

Then you need to solve

$$\int\frac{2t}{t^4-8t^2+1}dt.$$

This is routine work, by fraction decomposition.

$$\frac{\log(4 + \sqrt{15} - t^2) - \log(-4 + \sqrt{15} + t^2)}{2 \sqrt{15}}+C.$$

Answer 5

Substitute $x=\sqrt2\sinh t$:

$$\int\frac{\mathrm dx}{(x^2-3)\sqrt{x^2+2}}$$ $$=\int\frac{\mathrm dt}{2\sinh^2t-3}$$ $$=\int\frac{-\text{sech}^2t\mathrm dt}{3-5\tanh^2t}$$ $$=\frac1{2\sqrt{15}}\ln\left|\frac{\sqrt3+\sqrt5\tanh t}{\sqrt3-\sqrt5\tanh t}\right|+C$$ $$\therefore\boxed{\int\frac{\mathrm dx}{(x^2-3)\sqrt{x^2+2}}=\frac1{2\sqrt{15}}\ln\left|\frac{\sqrt5x-\sqrt3 \sqrt{2+x^2}}{\sqrt5x+\sqrt3 \sqrt{2+x^2}}\right|+C}$$

Answer 6

$\textbf{Case 1: }x\in(-\sqrt3,\sqrt3)\implies$Substitute $x=\sqrt3\tanh t$:

$$\int\frac{\mathrm dx}{(x^2-3)\sqrt{x^2+2}}$$ $$=-\frac1{\sqrt3}\int\frac{\mathrm dt}{\sqrt{3\tanh^2t+2}}$$ $$=-\frac1{\sqrt3}\int\frac{\cosh t\mathrm dt}{\sqrt{5\sinh^2t+2}}$$ $$=-\frac1{\sqrt{15}}\sinh^{-1}\left(\frac{\sqrt5\sinh t}{\sqrt2}\right)+C$$ $$=-\frac1{\sqrt{15}}\sinh^{-1}\left(\frac{\sqrt5|x|}{\sqrt2\sqrt{3-x^2}}\right)+C$$

$\textbf{Case 2: }x\in(-\infty,-\sqrt3)\cup(\sqrt3,\infty)\implies$Substitute $x=\sqrt3\coth t$:

$$\int\frac{\mathrm dx}{(x^2-3)\sqrt{x^2+2}}$$ $$=-\frac1{\sqrt3}\int\frac{\mathrm dt}{\sqrt{3\coth^2t+2}}$$ $$=-\frac1{\sqrt3}\int\frac{\sinh t\mathrm dt}{\sqrt{5\cosh^2t-2}}$$ $$=-\frac1{\sqrt{15}}\cosh^{-1}\left(\frac{\sqrt5\cosh t}{\sqrt2}\right)+C$$ $$=-\frac1{\sqrt{15}}\cosh^{-1}\left(\frac{\sqrt3\sqrt{x^2+2}}{\sqrt2\sqrt{3-x^2}}\right)+C$$