How do you integrate 2/((x-1)^2(x+1))2(x−1)2(x+1) using partial fractions?
1 Answer
int 2/((x-1)^2(x+1)) dx∫2(x−1)2(x+1)dx
= -1/2 ln abs(x-1)-1/(x-1)+1/2ln abs(x+1) + C=−12ln|x−1|−1x−1+12ln|x+1|+C
Explanation:
2/((x-1)^2(x+1)) = A/(x-1)+B/(x-1)^2+C/(x+1)2(x−1)2(x+1)=Ax−1+B(x−1)2+Cx+1
color(white)(2/((x-1)^2(x+1))) = (A(x-1)(x+1)+B(x+1)+C(x-1)^2)/((x-1)^2(x+1))2(x−1)2(x+1)=A(x−1)(x+1)+B(x+1)+C(x−1)2(x−1)2(x+1)
color(white)(2/((x-1)^2(x+1))) = ((A+C)x^2+(B-2C)x+(-A+B+C))/((x-1)^2(x+1))2(x−1)2(x+1)=(A+C)x2+(B−2C)x+(−A+B+C)(x−1)2(x+1)
Equating coefficients, we find:
{ (A+C=0), (B-2C=0), (-A+B+C=2) :}
Adding all three equations together, we find:
2B=2
and hence:
B=1
Then from the second equation we find:
C=1/2
Then from the first equation we find:
A=-1/2
So:
int 2/((x-1)^2(x+1)) dx
= int -1/2(1/(x-1))+1/(x-1)^2+1/2(1/(x+1)) dx
= -1/2 ln abs(x-1)-1/(x-1)+1/2ln abs(x+1) + C
