How do you integrate 6x2+1x2(x−1)2 using partial fractions?

1 Answer
Narad T.
Oct 22, 2016

The integral is ∫(6x2+1)dxx2(x−1)2=−1x+2lnx−2ln(x−1)−7x−1+C

Explanation:

Let's first determine the coefficients of the partial fractions

6x2+1x2(x−1)2=Ax2+Bx+Cx−1+D(x−1)2

6x2+1x2(x−1)2=A(x−1)2+Bx(x−1)2+Cx2(x−1)+Dx2x2(x−1)2

(6x2+1)=A(x−1)2+Bx(x−1)2+Cx2(x−1)+Dx2

Let x=0, then 1=A+0+0+0 ; So A=1
Let x=1, then 7=D ; so D=7
Also,coefficients of x3; 0=B+C, ; so B=−C
And coefficients of x2 ; 6=A−2B−C+D
So 6=1+2C−C+7 ⇒C=−2
And B=2

6x2+1x2(x−1)2=1x2+2x−2x−1+7(x−1)2

So

∫(6x2+1)dxx2(x−1)2=∫1dxx2+∫2dxx−∫2dxx−1+7dx(x−1)2

=−1x+2lnx−2ln(x−1)−7x−1+C

Related questions
See all questions in Integral by Partial Fractions
Impact of this question
2064 views around the world
You can reuse this answer
Creative Commons License