Difficulty in integrating I tried to integrate this by parts but it didn't work out. Any simple means of doing it.$$\int\sin^{-1}\biggl(\frac{2x+2}{\sqrt{4x^{2}+8x+13}}\biggr) \ dx$$

Put $2x+2 = 3 \tan(\theta)$ and see what happens.
$\textbf{Added.}$ First observe that $$4x^{2}+8x+13= (2x+2)^{2} + 3^{2}.$$ So I hope you are aware of the fact that $\text{if you have an integral of the form}$, $1+x^{2}$, then one generally substitutes, $x= \tan(\theta)$. That's the case here as well. By doing that we get, 
\begin{align*}
\int \sin^{-1}\biggl(\frac{2x+2}{\sqrt{4x^{2}+8x+13}}\biggr) \ dx &= \int\sin^{-1}\biggl(\frac{3\cdot \tan\theta}{3 \cdot \sec\theta}\biggr) \cdot 3 \sec^{2}\theta \ d\theta \\ &= 3\cdot\int \sin^{-1}(\sin\theta) \cdot \sec^{2}\theta \ d\theta \\ &= 3 \cdot \int \theta \cdot\sec^{2}\theta \ d \theta
\end{align*}
Use integration by parts to evaluate the last integral. Put $u = \theta$ and $dv = \sec^{2}(\theta) \ d\theta$.  So the answer for the last part should be $$\int \theta \cdot \sec^{2}\theta \ d \theta = \theta\cdot\tan\theta + \ln(\cos\theta) + C$$
