Originally Posted by
TheEmptySet When you did your u sub $\displaystyle u=x^{1/2}$
$\displaystyle \int \frac{x^{1/2}}{x-4}dx=2\int \frac{u^2}{u^2-4}du$
You had this, but
$\displaystyle 2\int \frac{u^2}{u^2-4}du=2\left( \int 1du +\int \frac{4}{u^2-4}du\right)$
Now we distribute the 2 to get
$\displaystyle 2\int du + 2\int \frac{4}{u^2-4}du$
pulling the four out in front of the 2nd integral gives
$\displaystyle 2\int du + (2)4\int \frac{1}{u^2-4}du$
factoring the denominator of the 2nd gives
$\displaystyle 2 \int du + 8 \int \frac{1}{(u-2)(u+2)}du$
Note that these are the same thing
$\displaystyle 8 \int \frac{1}{(u-2)(u+2)}du=8 \int \frac{du}{(u-2)(u+2)} $
I hope this clears it up.
Good luck.