# Integration partial fractions

• Feb 28th 2010, 07:22 AM
wolfhound
Integration partial fractions
Hello,
could someone tell me where I made a mistake on this partial fraction problem
• Feb 28th 2010, 07:32 AM
Quote:

Originally Posted by wolfhound
Hello,
could someone tell me where I made a mistake on this partial fraction problem

hi wolfhound,

$\displaystyle \frac{A}{x}+\frac{B}{x^2+1}=\frac{1}{x^3+x}$

$\displaystyle \frac{A\left(x^2+1\right)+Bx}{x\left(x^2+1\right)} =\frac{1}{x^3+x}$

$\displaystyle \frac{Ax^2+A+Bx}{x^3+x}=\frac{1}{x^3+x}$

$\displaystyle x\left(Ax+B\right)+A=x(0)+1$

$\displaystyle A=1$

$\displaystyle (1)x+B=0\ \Rightarrow\ B=-x$
• Feb 28th 2010, 07:39 AM
wolfhound
Hi archie
Oh so I dont actually get a value for B? except B=-x?
someone said I should input i in to solve these also
when should I take a coefficient as x and not a value please?
• Feb 28th 2010, 07:47 AM
Hi wolfhound,

sometimes there will be an x-component of the A and B values.
They are not necessarily "constants".

They are "numerators".

They only end up being constants sometimes.

If you add those two resulting fractions together,
you will see that the result is the original fraction you wanted to integrate.

since the sum of the resulting fractions is the original,
you can more conveniently integrate the partial fractions.

The trick is....

if the numerator of the original fraction is a constant,
then you can write it as "constant + (0)x"
• Feb 28th 2010, 07:55 AM
wolfhound
I see,
I am going to practice some more now
Thanks