Calculating the integral of a logarithmic expression. The problem I have been working with is 
$$\int \frac 1{\sqrt x(1+\sqrt x)}\,dx$$
The first step I did to solve this question was to set $u= 1+ \sqrt x$ the set $du = (1/2) x^{-1/2}$
Then I set $\dfrac 1 2\int\ln(du/u)$ and finally I think the answer is $\ln\dfrac 1 x+c$

Setting $u=1+\sqrt x$, we have $$\frac{du}{dx}=\frac{1}{2\sqrt x}\Rightarrow \frac{dx}{\sqrt x}=2du,$$ so
$$\int\frac{1}{1+\sqrt x}\cdot \frac{dx}{\sqrt x}=\int \frac 2u du=2\ln u+C=2\ln(1+\sqrt x)+C.$$
