# Thread: partial fractions - rationalizing substitutions

1. ## partial fractions - rationalizing substitutions

can someone help with the steps underlined, how do you i get from one to the other?

Thanks.

2. Originally Posted by calculus0
can someone help with the steps underlined, how do you i get from one to the other?

Thanks.
For this one..

$\frac{u^2}{u^2-4}=\frac{u^2-4+4}{u^2-4}=\frac{u^2-4}{u^2-4}+\frac{4}{u^2-4}=1+\frac{4}{u^2-4}$

3. Originally Posted by calculus0
can someone help with the steps underlined, how do you i get from one to the other?

Thanks.
$2 \int \frac{u^2}{u^2-4} = 2 \int \frac{u^2 -4 +4}{u^2-4} = 2 \int \frac{(u^2 -4) +4}{u^2-4}$

$= 2 \int [\frac{u^2 -4}{u^2-4} + \frac{4}{u^2-4}] = 2 \int [1+ \frac{4}{u^2-4}]$

4. so are you essentially adding and subtracting 4? (how would you know to do that?

5. Originally Posted by calculus0
so are you essentially adding and subtracting 4? (how would you know to do that?
When you have $\frac{u^2}{u^2-k}$

this can be rearranged to a form which can be directly integrated
using the tables of standard known integrals

by adding and subtracting k in the numerator, whatever constant k is.

$\frac{u^2}{u^2-k}=\frac{u^2-k+k}{u^2-k}=1+\frac{k}{u^2-k}$

both of which which can easily be integrated.

6. More generally, just do a "long division"

$u^2- 0u- 4)\overline{u^2+ 0u+ 0}$

Obviously the " $u^2$" will divide into the " $u^2$" 1 time so we will then need to subtract $(u^2+ 0u+ 0)- (u^2- 0u- 4)= 4$. $u^2- 4$ divides into $u^2$ 1 time with remainder 4: $\frac{u^2}{u^2- 4}= 1+ \frac{4}{u^2- 4}$.