Help with $\int \frac{x-\sqrt x}{x+\sqrt x} \, \mathrm d x$ I need help with this integral. I tried $$\int \frac{x-\sqrt x}{x+\sqrt x}\,\mathrm dx = \int \frac{(x-\sqrt x)^2}{x^2-x} \,\mathrm dx = \int \frac {x(x-2\sqrt x+1)}{x(x-1)}\,\mathrm dx = \int \frac {(x-2\sqrt x+1)}{(x-1)}\,\mathrm dx$$ but I don't know how to proceed? Thank you for your help. 

Before substitution:
$$\int \frac{x-\sqrt x}{x+\sqrt x} dx= \int \frac{\sqrt x -1}{\sqrt x+1} dx = \int \frac{\sqrt x +1-2}{\sqrt x+1} dx = \int \left(1- \frac{2}{\sqrt x+1} \right)dx$$ $$= x-2 \int \frac{1}{\sqrt x+1}dx$$
Then $x = (u-1)^2$:
$$\int \frac{1}{\sqrt x+1}dx = \int \frac{2(u-1)}{u} du = 2\int \left( 1- \frac{1}{u} \right) du$$
