1. Integration using Hyperbolic

$\displaystyle \int \frac{x}{\sqrt{x^2+4x}} dx$
use substitution $\displaystyle x+2=2cosh\theta$

$\displaystyle \int \frac{x}{\sqrt{(x+2)^2-4}} dx$

$\displaystyle \int \frac{x}{\sqrt{(2cosh\theta)^2-4}} dx$

$\displaystyle \int \frac{x}{\sqrt{(4cosh^2\theta)-4}} dx$

$\displaystyle \int \frac{x}{2\sqrt{(cosh^2\theta)-1}} dx$

$\displaystyle \int \frac{(2cosh\theta-2)2sinh\theta}{2sinh\theta} d\theta$

$\displaystyle \int 2cosh\theta-2 d\theta$

$\displaystyle = 2sinh\theta -2\theta + C$

but it doesnt seem right.

where did i go wrong?

thanks.

2. Seems good to me. But is difficult to get $\displaystyle \theta =f(x)$.
May be more simple is

$\displaystyle \int \frac{(x+2)-2}{\sqrt{(x+2)^2-4}} dx=\int \frac{x+2}{\sqrt{(x+2)^2-4}} dx-\int \frac{2}{\sqrt{(x+2)^2-4}} dx$

3. Looks wrong to me.
http://www.wolframalpha.com/input/?i=intg+%28x%29%2F%28x^2%2B4x%29^0.5+0+to+1
$\displaystyle \int \frac{x}{\sqrt{x^2+4x}} dx$
this gives 0.311

$\displaystyle \int 2cosh\theta-2 d\theta$
this gives 0.350

someone help?

thanks