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.
On
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.
$$\int \frac{12\sin^{-1} x}{(3x^2+1)^2}\,dx=\sin^{-1}x\int \frac{12x}{(3x^2+1)^2}\,dx -\int \left(\frac{1}{\sqrt{1-x^2}}\int \frac{12x}{(3x^2+1)^2}\,dx\right)\,dx$$ 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: $$\frac{-2\sin^{-1}x}{3x^2+1}+\tan^{-1}\left(\frac{2x}{\sqrt{1-x^2}}\right)+C$$