Q:

Integral of $ \frac{\sqrt{x+1}}{x} $.

Im looking for a way to calculate:
$$ \int{\frac{\sqrt{x+1}}{x}}  dx $$
I substituted $ u = \sqrt{x+1} $ and then $ dx = 2 * \sqrt{x+1}  du $ so I have:
$$ \int{\frac{u}{x} * 2 \sqrt{x+1} du } $$
$ x = u^2-1 $ 
and
$$ 2 \int{\frac{u^2}{u^2-1} } du  = 2 \int{\frac{u^2}{(u-1)(u+1)} } du $$
I'm not sure if this is correct. Maybe I had some errors on the way here.
But if this is correct. How would I proceed then?

A:

What you did was correct. To finish it off, notice that
$$\dfrac{u^2}{u^2-1} = \dfrac{(u^2-1)+1}{u^2-1} = 1 + \dfrac{1}{u^2-1}$$
The integral can then be computed using partial fractions:
$$\dfrac{1}{u^2-1} = \dfrac{1}{(u-1)(u+1)} = \dfrac{1}{2} \left( \dfrac{1}{u-1} - \dfrac{1}{u+1} \right)$$

