<s>[INST] 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:

 [/INST] 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!

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