A question about Indefinite trigonometric integration I am practising a unit on Integration. I am going through some past
year papers, and there are  some types of questions that I could not
solve. So if anyone could help me in this, I'd really appreciate this.

Evaluate the following integrals:
  $$
i)\quad\int\left(\frac{2}{\sqrt{x}}+2e^{-4x}+\frac{1}{3(1-x)}\right)\:\mathrm{d}x
$$

In this, I managed to get to a point where the answer is:
$$
4\sqrt{x}-\frac{e^{-4x}}{2}+\frac{\ln|x+1|}{3}+C
$$
Is this the final answer or there's more I can do here. I'm specially confused about the $|x+1|$ part.
Next is a trigonometric substituition question. I've tried basic ones of this type, but this one is very difficult and complicated for me. If someone could point me into a direction then maybe I can try solving.

Show that:
  $$
\int\frac{x^2}{\sqrt{36-x^2}}\:\mathrm{d}x=18\sin^{-1}\left(\frac{x}{6}\right)-\frac{1}{2}x\sqrt{36-x^2}+C
$$
  with an appropriate trigonometric substitution.

 A: *

*i)


The result has to be slightly modified; the last function is $-\frac{1}{3}\ln|1-x|$; in fact its derivative w.r.t. $x$ is equal to
$$-\frac{1}{3}\frac{-1}{1-x}=\frac{1}{3(1-x)}, $$
if  $1-x>0$ and
$$-\frac{1}{3}\frac{1}{x-1}=\frac{1}{3(1-x)}, $$
if $1-x<0$. 
Moreover, the result of the integral is given by the sum of the proposed functions (with the replacement of the third with the one above) plus an arbitrary constant $C$.


*

*ii)


With the substitution $x=6\sin t$, $dx=6\cos t dt$, the proposed integral is equal to
$$\int \frac{36\sin^2 t}{6\sqrt{1-\sin^2 t}}6\cos t dt=\int \frac{36\sin^2 t}{|\cos t|}\cos t dt= \int36\sin^2 t dt,  $$
as $\cos t> 0$, if  $t\in(-\frac{\pi}{2},\frac{\pi}{2})$, which is the taken image of $\operatorname{arcsin}(\cdot)$.
All you need is
$$36\int\sin^2 t dt=36\left(\frac{1}{2}t-\frac{1}{4}\sin(2t)\right)+C,$$
with $\sin(2t)=2\sin t\cos t$.
A: Here we can put:
$$ 
x=6\sin\theta,\:\text{then we can get}\;\frac{\mathrm{d}x}{\mathrm{d}\theta}=6\cos\theta\:\mathrm{d}\theta
$$
so that your integral becomes: 
$$
\int\frac{36\sin^2\theta(6\cos\theta)}{6\cos\theta}\:\mathrm{d}\theta=\int 36\sin^2\theta\:\mathrm{d}\theta
$$
I think, from this you should be able to proceed further.
EDIT: For your first question, yes that is the final answer except the fact that you made a small mistake. Always to check your answer, differentiate it and see if it's the same as your question.
ANS:
$$
4\sqrt{x}-\frac{e^{-4x}}{2}+\frac{\ln|1-x|}{3}+C
$$
