Originally Posted by
larson For the first one:
$\displaystyle \int x^2 + \frac{1}{(3x)^2}$
I would just split them up to make it easier like this and also just fix up that $\displaystyle 3x^2$ to make it look a bit nicer:
$\displaystyle \int x^2 + \int \frac{1}{9x^2}$
The first part is easy...
$\displaystyle \int x^2 = \frac {x^3}{3}$
The second part is a little bit harder...
$\displaystyle \int \frac{1}{9x^2} = \frac{1}{9} \int x^-2$
Then for the $\displaystyle \int x^-2$ all it is... is $\displaystyle \frac {x^-1}{-1}$
So for a final answer you have...
$\displaystyle \frac {x^3}{3} - \frac{1}{9x}$