# Thread: partial fractions decomposition problem

1. ## partial fractions decomposition problem

can someone solve

∫ (4dx)/(x²-2x-3)

i don't think this equation diverges i think the denominator is 10 and not 4 but i am not sure

2. Originally Posted by cummings15
can someone solve

∫ (4dx)/(x²-2x-3)

i don't think this equation diverges i think the denominator is 10 and not 4 but i am not sure
$\displaystyle \frac{4}{(x+1)(x-3)} = \frac{A}{x+1} + \frac{B}{x-3}$

$\displaystyle 4 = A(x-3) + B(x+1)$

let $\displaystyle x = 3$ ... $\displaystyle B = 1$

let $\displaystyle x = -1$ ... $\displaystyle A = -1$

$\displaystyle \frac{4}{(x+1)(x-3)} = \frac{1}{x-3} -\frac{1}{x+1}$

now you can integrate the right side ...

3. Originally Posted by cummings15
can someone solve

∫ (4dx)/(x²-2x-3)

i don't think this equation diverges i think the denominator is 10 and not 4 but i am not sure
$\displaystyle \frac{4}{x^2-2x-3} = \frac{4}{(x+1)(x-3)}$

Now find $\displaystyle A$ and $\displaystyle B$ such that $\displaystyle \frac{4}{(x+1)(x-3)} =\frac{A}{x+1} +\frac{B}{x-3}$

4. would it be

4ln l(x-3)(x+1)l +c

5. No. Skeeter had already worked out that the coefficients A and B are 1 and -1 so the integral is 4(ln|x-3|- ln|x+1|)+ C.

6. so then i can distribute and get

4ln l(x-3)(x+1)l + C