Difficulty in integrating I tried to integrate this by parts but it didn't work out. Any simple means of doing it.$$\int\sin^{-1}\biggl(\frac{2x+2}{\sqrt{4x^{2}+8x+13}}\biggr) \ dx$$
 A: 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 $$\int \theta \cdot \sec^{2}\theta \ d \theta = \theta\cdot\tan\theta + \ln(\cos\theta) + C$$
A: Let's not start to integrate in too much of a hurry.
Completing the square is a natural move: $4x^2+8x+13=(2x+2)^2+9$.
In a right triangle, a certain angle has sine equal to
$$\frac{2x+2}{\sqrt{(2x+2)^2 +9}}.$$
If the "opposite" side is $2x+2$ and the hypotenuse is $\sqrt{(2x+2)^2+9}$, the remaining side of the triangle must be $3$. 
So we are trying to integrate
$$\arctan\left(\frac{2x+2}{3}\right).$$
After the natural substitution, we arrive at
$$\int\frac{3}{2}\arctan t \,dt,$$
a mild variant of a standard integral.  
Added for completeness:  We find $\int \arctan t\;dt$, using integration by parts. Let $u=\arctan t$, $dv=dt$. Then $du=\frac{dt}{1+t^2}$ and $v$ can be taken to be $t$.  It follows that 
$$\int \arctan t\;dt=t\arctan t -\int \frac{t}{1+t^2} dt.$$
The remaining integral yields easily to the substitution $w=1+t^2$.
Finally, we remember about our constant factor $3/2$ and obtain
$$\frac{3}{2}t \arctan t -\frac{3}{4}\ln(1+t^2) + C.$$
We omit the back substitution $t=(2x+2)/3$.
The issue of signs: What about if $2x+3$ is negative?  For indefinite integrals, there is almost a calculus tradition not to worry about such things. But let's.  The usual meaning of  $\arcsin u$ is the number between $-\pi/2$ and $\pi/2$ whose sine is $u$. For $\arctan u$, there are two common choices of interval,  $(-\pi/2,\pi/2)$ and $[0,\pi)$. If we use $(-\pi/2,\pi/2)$, the answer obtained above is correct for negative $2x+3$.  If we use $[0,\pi)$, a small adjustment needs to be made.
