How do you use partial fraction decomposition to decompose the fraction to integrate $(x^2+x+1)/(1-x^2)$ ?

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Answer 1 · 2015-07-27T16:24:09

Divide first, then find the partial fraction decomposition.
$(x^2+x+1)/(1-x^2) = -1 - (3/2)/(x-1) + (1/2)/(x+1)$

Before looking for a partial fraction decomposition, we must have the degree of the denominator strictly less than that of the numerator.

So we need to divide or regroup to get:

$(x^2+x+1)/(1-x^2) = (x^2-1+x+2)/(1-x^2)$

$= (x^2-1)/(1-x^2)+(x+2)/(1-x^2)$

$= -1 - (x+2)/(x^2-1)$

Now we can get the partial fraction decomposition for:

$(x+2)/(x^2-1) = (x+2)/((x-1)(x+1)) = A/(x-1)+B/(x+1)$

We need $Ax+A+Bx-B = x+2$

So

$A+B = 1$
$A-B = 2$

$2A = 3$ so $A = 3/2$ and $B = -1/2$

Putting it all together we have:

$(x^2+x+1)/(1-x^2) = -1-((3/2)/(x-1) - (1/2)/(x+1))$

$= -1 - (3/2)/(x-1) + (1/2)/(x+1)$

Now integrate.

How do you use partial fraction decomposition to decompose the fraction to integrate (x^2+x+1)/(1-x^2)(x^2+x+1)/(1-x^2)?