How do you integrate ∫x2+5x−7x2(x+1)2 using partial fractions?

1 Answer
Narad T.
Feb 3, 2017

The answer is =7x+19ln(|x|)+11x+1−19ln(|x+1|)+C

Explanation:

Let's perform the decomposition into partial fractions

x2+5x−7x2(x+1)2=Ax2+Bx+C(x+1)2+Dx+1

=A(x+1)2+Bx(x+1)2+Cx2+Dx2(x+1)(x2(x+1)2)

As the denominators are the same, we can compare the numerators

x2+5x−7=A(x+1)2+Bx(x+1)2+Cx2+Dx2(x+1)

Let x=0, ⇒, −7=A

Let x=−1, ⇒, −11=C

Coefficients of x3, ⇒, 0=B+D

Coefficients of x, ⇒, 5=2A+B

B=5−2A=5+14=19

D=−B=−19

Therefore,

x2+5x−7x2(x+1)2=−7x2+19x−11(x+1)2−19x+1

So,

∫(x2+5x−dxx2(x+1)2=−7∫dxx2+19∫dxx−11∫dx(x+1)2−19∫dxx+1

=7x+19ln(|x|)+11x+1−19ln(|x+1|)+C

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