# I need help with partial fractions

• Sep 24th 2009, 10:03 PM
farisallil
I need help with partial fractions
I need to compute this integral:
int ((1-x^2)/(x^3+x)) dx
I think I should use the partial fraction method to simplify the fraction
so
(1-x^2)/(x^3+x)= A/x + B/ (1+x^2)
Therefore
A(1+x^2)+B(x)=1-x^2
putting it in a polynomial form:
x^2 (A)+x (B) + 1 (A)= 1- x^2
and by equating the coefficients,
A = -1
B = 0
However, A doesn't satisfy the constant on the LHS (+1)

So what is the thing that I did wrong??

• Sep 25th 2009, 05:05 AM
Amer
Quote:

Originally Posted by farisallil
I need to compute this integral:
int ((1-x^2)/(x^3+x)) dx
I think I should use the partial fraction method to simplify the fraction
so
(1-x^2)/(x^3+x)= A/x + B/ (1+x^2)
Therefore
A(1+x^2)+B(x)=1-x^2
putting it in a polynomial form:
x^2 (A)+x (B) + 1 (A)= 1- x^2
and by equating the coefficients,
A = -1
B = 0
However, A doesn't satisfy the constant on the LHS (+1)

So what is the thing that I did wrong??

ok

$\frac{1-x^2}{x^3+x}=\frac{1-x^2}{x(x^2+1)}$

if you have a denominator with a second degree the nominator of it should be a first degree not constant so

$\frac{1-x^2}{x^3+x} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$

to find A,B and C

$1-x^2 = A(x^2+1) + Bx^2 + Cx$

$A+B = -1$
$C =0$
$A=1$ find B from the last one

so you have A=1 , B=-2 ,C=0

$\frac{1-x^2 }{x^3 + x} = \frac{1}{x} + \frac{-2x}{x^2+1}$

to make sure my solution is correct

$\frac{1}{x} + \frac{-2x}{x^2+1} = \frac{x^2+1}{x^3+x} + \frac{-2x^2}{x^3+x } =\frac{1-x^2 }{x^3+x}$
and this is the left side
• Sep 25th 2009, 05:08 AM
farisallil
ooooooo (Surprised)
I see

Thank you very much Amer, I really appreciate your help (Nod)