Hi there, I'm facing this problem which I couldn't solve.

The statement says:

Evaluate the next integrals, expressing it previously in the forms that contains $\displaystyle \sqrt[ ]{a^2+u^2}$, $\displaystyle \sqrt[ ]{a^2-u^2}$ and $\displaystyle \sqrt[ ]{u^2-a^2}$, and then solve it using the integral table;

The exercise which I couldn't solve is:

$\displaystyle \displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sq rt[ ]{x^2+2x}}dx$

I started by completing the square:

$\displaystyle x^2+2x=(x+1)^2-1$

Then

$\displaystyle u^2=(x+1)^2$

$\displaystyle u=(x+1)\Rightarrow{x+3=u+2}$

$\displaystyle du=dx$

$\displaystyle \displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sq rt[ ]{x^2+2x}}dx=\displaystyle\int_{}^{}\displaystyle\f rac{u+2}{\sqrt[ ]{u^2-1}}du$

From here I've tried to solve it by parts

$\displaystyle t=u+2$

$\displaystyle dt=du$

$\displaystyle dv=\displaystyle\frac{1}{\sqrt[ ]{u^2-1}}du$

$\displaystyle v=\ln|u+\sqrt[ ]{u^2-1}|$

$\displaystyle \displaystyle\int_{}^{}\displaystyle\frac{u+2}{\sq rt[ ]{u^2-1}}du=(u+2)\ln|u+\sqrt[ ]{u^2-1}|-\displaystyle\int_{}^{}\ln|u+\sqrt[ ]{u^2-1}|du$

If I try again by parts with $\displaystyle -\displaystyle\int_{}^{}\ln|u+\sqrt[ ]{u^2-1}|du$ it gets more complicated.

So, what do you say?

Bye there.