How do you use partial fractions to find the integral ∫4x2x3+x2−x−1dx?

1 Answer
Andrea S.
Dec 5, 2016

∫4x2dxx3+x2−x−1=ln|x−1|+3ln|x+1|+2x+1

Explanation:

The general method to integrate proper rational functions is to develop them in a sum of partial fractions.

In order to do that, we must first factorize the denominator.
You can easily see (x+1) and (x-1) are factors and write:

(x3+x2−x−1)=(x+1)(x2−1)=(x+1)2(x−1)

The decomposition in partial fraction is thus:

4x2x3+x2−x−1=Ax−1+Bx+1+C(x+1)2

4x2(x+1)2(x−1)=A(x+1)2+B(x−1)(x+1)+C(x−1)(x+1)2(x−1)

Ax2+2Ax+A+Bx2−B+Cx−C=4x2

Equating the coefficientS of the same grade in x we have the linear system:

A+B=4
2A+C=0
A−B−C=0

that can be solved as:

A=1,B=3,C=−2

So:

∫4x2dxx3+x2−x−1=∫dxx−1+∫3dxx+1−∫2dx(x+1)2

∫4x2dxx3+x2−x−1=ln|x−1|+3ln|x+1|+2x+1

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