1. ## integration help

right why does this happen? can someone explain what is going on?

2. 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 split it into two easier integrals: $\displaystyle \frac{1}{2} \int \frac{1}{x^2}~dx - \frac{1}{2} \int \frac{1}{x^2+2}~dx$

First integral should be simple enough for you, and as for the second integral, you should see that by taking 2 out as a factor and doing some manipulation, you'll get a standard arctan form.

3. 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.
I am having great difficulty understanding your formulas- can you write them down traditionally?

EDIT: nvm now they all of a sudden look good

4. ln x does not apply here since the differential of the denominator does not appear at the top.

The method that you have posted appears to be a partial fraction method.