I need to find the Integral of Sqrt(x^2+2x)dx. It is a Integration by Partial Fractions exercise.
I don't even know where to start. Any help on where to start would be greatly appreciated.
Not a typo there neither. It's the integral of the square root of (x^2 + 2x) dx
Krizalid: If it isn't necessary I don't have to use trigonometric expressions. If it's the only way, please explain a bit further.
EDIT: If this helps, the solution given by the book is:
(1/2)[(x+1)Sqrt(x^2+2x)-ln|x+1+Sqrt(x^2+2x)|]+c
I know something was wrong.
Well then, to perform a trig. substitution, we have $\displaystyle x^2+2x=x^2+2x+1-1=(x+1)^2-1$ so you need to put $\displaystyle x+1=\sec\varphi.$
(I closed your duplicate thread on calculus section, there's no reason to make an identical one.)
$\displaystyle \int\sec^3z\,dz=\int\sec z\sec^2z\,dz=\int\sec z(\tan z)'\,dz.$
Strongly recommended lecture: make your partial integration faster. (See my signature.)
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Or if you want to use the classic way, put $\displaystyle u=\sec z$ and $\displaystyle dv=\sec^2z\,dz.$