1. Evaluate the indefinite integral

I can't seem to figure out how to integrate

2x
---- dx ?
x-4

I tried U substitution for both 2x and x-4 but to no avail

2. Originally Posted by nottheface
I can't seem to figure out how to integrate

2x
---- dx ?
x-4

I tried U substitution for both 2x and x-4 but to no avail
Hint:

$\dfrac{2x}{x-4}=\dfrac{2x-8+8}{x-4}=\dfrac{2(x-4)+8}{x-4}=2+\dfrac{8}{x-4}$

3. Another way to do the same thing: substitute u= x- 4 as you did. Then x= u+ 4 so 2x= 2u+ 8. du= dx and the integral becomes $\int \frac{2u+ 8}{u}du= \int \left(2+ \frac{8}{u}\right) du$