Originally Posted by
Chop Suey First, you used 2 to multiply by 1: $\displaystyle \frac{1}{2} \int \frac{2}{x^2(x^2+2)}$
Then, you used $\displaystyle x^2$ to add 0: $\displaystyle \frac{1}{2} \int \frac{(2+x^2)-x^2}{x^2(x^2+2)}$
Then, you split $\displaystyle (2+x^2)$ and $\displaystyle x^2$: $\displaystyle \frac{1}{2} \int \frac{2+x^2}{x^2(2+x^2)}~dx - \frac{1}{2} \int \frac{x^2}{x^2(x^2+2)}~dx$
And as you can see in the end, all this work was an elaborate plan to achieve a cancellation and get a more simpler integral: $\displaystyle \frac{1}{2} \int \frac{1}{x^2}~dx - \frac{1}{2} \int \frac{1}{x^2+2}~dx$
First integral should be simple enough for you, and as for the second integral, you should see that by taking 2 common factor and doing some manipulation, you'll get a standard arctan form.