Integration of $\arcsin(f(x))$ $$\int\sin^{-1}\left(\dfrac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx$$
$\text {attempt}$ Let $x+1=u$, then the internal function becomes $\frac {2u}{\sqrt{4u^2+9}}$, and $dx=du$. Then I attempted by parts but derivative of $sin^{-1}(f(x))$ is troubling me in second part of IBP. Any better approach, or am I missing on a trick? I only know that the answer contains $\ln$ and $\tan^{-1}$. Thanks!
 A: Let $$I = \int \sin^{-1}\left(\frac{2x+2}{\sqrt{(2x+2)^2+3^3}}\right)dx$$
Now Put $2x+2=3\tan \theta\;,$ Then $\displaystyle dx = \frac{3}{2}\sec^2 \theta d \theta$
So Using Integration by parts $$I = \frac{3}{2}\int \theta \sec^2 \theta d \theta = \frac{3}{2}\left[\theta \tan \theta+\ln |\sec \theta|\right]+\mathcal{C}$$
So $$I = \frac{3}{2}\cdot \frac{2x+2}{3}\tan^{-1}\left(\frac{2x+2}{3}\right)+\frac{3}{2}\ln \left(\frac{\sqrt{3^2+(2x+2)^2}}{3}\right)+\mathcal{C}$$
So $$I = (x+1)\cdot \tan^{-1}\left(\frac{2x+2}{3}\right)+\frac{3}{2}\ln \left(\frac{\sqrt{3^2+(2x+2)^2}}{3}\right)+\mathcal{C}$$
A: Let us use integration by parts 
$$u=\sin ^{-1}\left(\frac{2 x+2}{\sqrt{4 x^2+8 x+13}}\right)\implies u'=\frac 6 {4 x^2+8 x+13}$$ and $v=x$. So
$$I=\int \sin ^{-1}\left(\frac{2 x+2}{\sqrt{4 x^2+8 x+13}}\right)\,dx=x\sin ^{-1}\left(\frac{2 x+2}{\sqrt{4 x^2+8 x+13}}\right)-\int \frac{6x}{4 x^2+8 x+13}\,dx$$ Now, using the roots of the quadatic in denominator an partial fractions $$\frac{6x}{4 x^2+8 x+13}=\frac{\frac{3}{2}-i}{2 x+(2+3 i)}+\frac{\frac{3}{2}+i}{2 x+(2-3 i)}$$ Integrating is now easy : you get two terms $a_i \log(b_i)$ in which $a_i,b_i$ are complex numbers. Expanding, you should find a logarithm and an arctangent.
Otherwise, write $$\frac{6x}{4 x^2+8 x+13}=\frac{6x+6-6 }{4 x^2+8 x+13 }=\frac 34\frac{8x+8-8 }{4 x^2+8 x+13 }$$ $$\frac{6x}{4 x^2+8 x+13}=\frac 34 \frac{(4 x^2+8 x+13)'}{4 x^2+8 x+13 }-\frac 6 {4 x^2+8 x+13 }$$ For the last piece, complete the square and things will become quite obvious.
