Compute : $\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$ Question: Compute this integral 
$$\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$$
My Approach: 
$$\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$$
$$=\int\frac{x+2}{\sqrt{x^2+5x}+6}\times \frac{{\sqrt{x^2+5x}-6}}{{\sqrt{x^2+5x}-6}}~dx$$
$$\int\frac{(x+2)(\sqrt{x^2+5x})}{x^2+5x-36}~dx~~- \underbrace {~\int\frac{(6x+12)}{x^2+5x-36}~dx~}_{\text{This one I know how to deal with} }$$
$$\text{Now:} ~\int\frac{(x+2)(\sqrt{x^2+5x})}{x^2+5x-36}~dx$$
$$=\frac{1}{2}\int\frac{(2x+5-1)(\sqrt{x^2+5x})}{x^2+5x-36}~dx$$
$$=\frac{1}{2}\int\frac{(2x+5)(\sqrt{x^2+5x})}{x^2+5x-36}~dx~~- \frac{1}{2}\int\frac{(\sqrt{x^2+5x})}{x^2+5x-36}~dx$$
$$\Big( \text{Let} ~ x^2+5x=t \implies (2x+5)~dx = dt \Big)$$
$$ \underbrace{\frac{1}{2}\int \frac{\sqrt{t}}{t-36}~dt}_{\text{I can deal with this}} ~~- \frac{1}{2}\int \frac{\sqrt{x^2+5x}}{x^2+5x-36}~dx$$
Now I'm stuck. I am unable to calculate: $$ \int \frac{\sqrt{x^2+5x}}{x^2+5x-36}~dx$$
P.S.: I am high school student so please try to use elementary integrals only; i.e. integration by parts and substitution. I don't know how to use complex numbers in integration, multiple integrals, error function, etc. (I don't know if it can be used here or not, just clarifying.) 
As answered by @Kanwaljit Singh: Finally I have to compute:
$$\int \frac{1}{\sqrt{x^2+5x}-6}$$
But if I was able to compute it, I would have done it in the very first step, id est ;
$$\int \frac{x+2}{\sqrt{x^2+5x}+6}~dx = 
\frac{1}{2}\int \frac{2x+5-1}{\sqrt{x^2+5x}+6}~dx
\\ \frac{1}{2}\int \frac{2x+5}{\sqrt{x^2+5x}+6}~dx ~- \frac{1}{2}\int \frac{1}{\sqrt{x^2+5x}+6}~dx
\\ \Big( \text{Let} ~ x^2+5x=t \implies (2x+5)~dx = dt \Big)
\\ \underbrace{\frac{1}{2}\int \frac{1}{t+6}~dt}_{\text{Doable}} ~-~\frac{1}{2}\int \frac{1}{\sqrt{x^2+5x}+6}~dx
\\ \int \frac{1}{\sqrt{x^2+5x}+6}~dx $$
Reached to a similar step by a short path.
But how do I compute this one?
A screenshot of this question:

 A: $$ \int \frac{\sqrt{x^2+5x}}{x^2+5x-36}~dx$$
$$=\int \frac{\sqrt{x^2+5x}+6-6}{(\sqrt{x^2+5x})^2-6^2}~dx$$
$=\int \frac{\sqrt{x^2+5x}+6}{(\sqrt{x^2+5x}+6)(\sqrt{x^2+5x}-6)} - \int \frac{6}{x^2+5x-36} ~dx$
$$=\int \frac{1}{\sqrt{x^2+5x}-6} - \int \frac{6}{x^2+5x-36} ~dx$$
Hope you can proceed further.
A: Introduce the Euler substitution:
Let $u=\dfrac{\sqrt{x^2+5x}}{x}$ ,
Then $x=\dfrac{5}{u^2-1}$
$dx=-\dfrac{10u}{(u^2-1)^2}~du$
$\therefore\int\dfrac{x+2}{\sqrt{x^2+5x}+6}~dx$
$=-\int\dfrac{\dfrac{5}{u^2-1}+2}{\dfrac{5u}{u^2-1}+6}\dfrac{10u}{(u^2-1)^2}~du$
$=-\int\dfrac{(2u^2+3)10u}{(6u^2+5u-6)(u^2-1)^2}~du$
$=-\int\dfrac{(2u^2+3)10u}{(3u-2)(2u+3)(u+1)^2(u-1)^2}~du$
$=\int\left(-\dfrac{5}{2(u+1)^2}-\dfrac{5}{2(u-1)^2}+\dfrac{11}{2(u+1)}+\dfrac{13}{2(u-1)}-\dfrac{144}{13(2u+3)}-\dfrac{252}{13(3u-2)}\right)~du$ (according to http://www.wolframalpha.com/input/?i=-((2u%5E2%2B3)10u)%2F((3u-2)(2u%2B3)(u%2B1)%5E2(u-1)%5E2))
$=\dfrac{5}{2(u+1)}+\dfrac{5}{2(u-1)}+\dfrac{11\ln(u+1)}{2}+\dfrac{13\ln(u-1)}{2}-\dfrac{72\ln(2u+3)}{13}-\dfrac{84\ln(3u-2)}{13}+C$
$=\dfrac{5u}{u^2-1}+\dfrac{11\ln(u+1)}{2}+\dfrac{13\ln(u-1)}{2}-\dfrac{72\ln(2u+3)}{13}-\dfrac{84\ln(3u-2)}{13}+C$
$=\sqrt{x^2+5x}+\dfrac{11}{2}\ln\dfrac{x+\sqrt{x^2+5x}}{x}+\dfrac{13}{2}\ln\dfrac{\sqrt{x^2+5x}-x}{x}-\dfrac{72}{13}\ln\dfrac{3x+2\sqrt{x^2+5x}}{x}-\dfrac{84}{13}\ln\dfrac{3\sqrt{x^2+5x}-2x}{x}+C$
A: all of these answers are so stupendously complicated yo! here's a much better method:
as stated, we need to deal with integral of 1/(sqrt(x^2+5x)+6)
write x^2+5x= (x+5/2)^2 - (5/2)^2
now substitute x+5/2 = 5y/2
now, our main focus after taking the constants out becomes the integral of 
1/(sqrt(y^2 - 1) + 12/5)
put arcsecy= z or y= secz so integral after getting rid of constants is
secz.tanz/(tanz + a) where a = 12/5.
now, here comes the genius step:
secz.tanz/(tanz + a)= secz - asecz/(tanz + a). else this prob was extremely hard. Obviously we already know integral of secz.
For secz/(tanz + a), write it as 1/(sinz + acosz). now this is the standard form that can be solved by either of the 2 methods:


*

*Let 1= rcosq and a= rsinq and hence forming a sine term in the denominator.

*Or even better, go for u= tan(z/2) substitution (also referred to as the universal substitution) and you will get an easy standard rational function that even a 4 yo can integrate!

