# Thread: Partial fraction

1. ## Partial fraction

$\int \frac {x^2}{(x-3)(x+2)^2} ~dx = \frac {A}{x-3} + \frac {B}{x+2} + \frac {C}{(x+2)^2}$

That is where I am at so far, and I just need a little help with finding the common denominator. Normally you would multiply A by the denominators of both B and C, but what about in this case? Since they are practically the same. Would I multiply A by $(x+2)(x+2)^2$ or just by (x+2)?

2. Originally Posted by redman223
$\int \frac {x^2}{(x-3)(x+2)^2} ~dx = \frac {A}{x-3} + \frac {B}{x-2} + \frac {C}{(x-2)^2}$

That is where I am at so far, and I just need a little help with finding the common denominator. Normally you would multiply A by the denominators of both B and C, but what about in this case? Since they are practically the same. Would I multiply A by $(x-2)(x-2)^2$ or just by (x-2)?
the common denominator is always the denominator of the original fraction. are you sure it should be x - 2 or x + 2?

3. Oops, it was supposed to be x+2, it has been fixed. So how are those treated?

4. Hello,

Multiply the whole by $(x+2)^2$, but by $(x-3)(x+2)^2$ would be more straightforward :

$x^2=A(x+2)^2+B(x+2)(x-3)+C(x-3)$

Then, let $x=-2$ to annulate x+2 ---> you will have C
After that, let $x=3$ to annulate x-3 ---> you will have A
And then, pick up a random value for x, for example $x=0$, to get B.

5. How does setting x = 0 allow you to find B?

6. ## See the basics

Originally Posted by redman223
How does setting x = 0 allow you to find B?
Actually didn't understood what you actually meant.
Suppose
(x+1)/(x+2)(x+3)=[A/(x+2)] +B/x+3 or we may say
(x+1)/(x+2)(x+3)=[A(x+3)+B(x+2)]/(x+2)(x+3) or
x+1=A(x+3)+B(x+2)
if you put x=-2 you will get
-1=A(-2+3)+B(-2+2) or
-1=A
or A=-1
we take such value of x for which atleast 1 expression may become 0 so that calculating other constant may be calculated easily.
NOTE:actually same thing can be done by comparing LHS and RHS. But this substitution method is much easier.

7. Originally Posted by redman223
How does setting x = 0 allow you to find B?
Once you find the value for $A$ and $C$, you will only have $B$ as an unknown in the equation (left to found) when $x=0$ or any value you choose because you can insert values for $A$ and $C$.