<s>[INST] 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!
 [/INST] 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.
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