$\displaystyle \int \frac{sin(2x) - x}{x^2 + cos(2x)} dx$
No clue how to start this problem. Any help appreciated, I'll be looking through my notes to see if we've done something similar to this one in the mean time.
Ok here's my work:
$\displaystyle u=x^2+Cos(2x)$
$\displaystyle du=3x-2Sin(2x)dx \rightarrow du=-2(-x+Sin(2x))dx\rightarrow -\frac{1}{2}du=-x+Sin(2x)dx$
$\displaystyle -\frac{1}{2}\int \frac{1}{u}du \rightarrow -\frac{1}{2}[ln|u|]$
$\displaystyle -\frac{1}{2}ln|x^2+Cos(2x)|+C$
Hows it look?