Prove $\int\frac{12x\sin^{-1}x}{9x^4+6x^2+1}dx=-\frac{2\sin^{-1}x}{3x^2+1}+\tan^{-1}\left(\frac{2x}{\sqrt{1-x^2}}\right)+C$ How to prove
$$\int\frac{12x\sin^{-1}x}{9x^4+6x^2+1}dx=-\frac{2\sin^{-1}x}{3x^2+1}+\tan^{-1}\left(\frac{2x}{\sqrt{1-x^2}}\right)+C$$
where $\sin^{-1}x$ and $\tan^{-1}x$ are inverse of trig functions. I don't know how to find the integral because of inverse of trig functions. I missed calc class twice. Please help me. Thanks.
 A: $$\int \frac{12\sin^{-1} x}{(3x^2+1)^2}\,dx=\sin^{-1}x\int \frac{12x}{(3x^2+1)^2}\,dx -\int \left(\frac{1}{\sqrt{1-x^2}}\int \frac{12x}{(3x^2+1)^2}\,dx\right)\,dx$$
To evaluate $\displaystyle \int \frac{12x}{(3x^2+1)^2}$, use the substitution $3x^2+1=u \Rightarrow 6x\,dx=du$ to get: $\frac{-2}{3x^2+1}$, i.e
$$\int \frac{12\sin^{-1} x}{(3x^2+1)^2}\,dx=\frac{-2\sin^{-1}x}{3x^2+1}+2\int \frac{1}{\sqrt{1-x^2}(3x^2+1)}\,dx$$
To evaluate the last integral, use the substitution $x=\sin\theta \Rightarrow dx=\cos\theta d\,\theta$ to get:
$$2\int \frac{1}{\sqrt{1-x^2}(3x^2+1)}\,dx=2\int \frac{1}{3\sin^2\theta+1}\,d\theta=2\int \frac{\sec^2\theta}{4\tan^2\theta+1}\,d\theta$$
$$\Rightarrow 2\int \frac{\sec^2\theta}{4\tan^2\theta+1}\,d\theta=\tan^{-1}(2\tan\theta)+C$$
Since $\sin\theta=x$, it is easy to see that $\tan\theta=x/\sqrt{1-x^2}$, hence
$$\tan^{-1}(2\tan\theta)=\tan^{-1}\left(\frac{2x}{\sqrt{1-x^2}}\right)+C$$
Putting everything together, the final answer is:
$$\frac{-2\sin^{-1}x}{3x^2+1}+\tan^{-1}\left(\frac{2x}{\sqrt{1-x^2}}\right)+C$$
A: Hint
As said in comments, integration by parts is the only way to get rid of the inverse trigonometric functions.
So, let $$u=\sin ^{-1}(x)$$ $$v'=\frac{12x}{9x^4+6x^2+1}=\frac{12x}{(3x^2+1)^2}$$ Then $$u'=\frac{1}{\sqrt{1-x^2}}$$ $$v=-\frac{2}{3 x^2+1}$$ Now, the answer which is given to you suggests the change of variable to be used.
I am sure that you can take from here.
