How do you integrate ∫(dxx(x+1)2) using partial fractions?
1 Answer
Feb 8, 2016
Explanation:
As stated by the question, we must first express
1x(x+1)2≡Ax+Bx+1+C(x+1)2 ,
where
Which means,
1≡A(x+1)2+Bx(x+1)+Cx .
Simplifying gives
(A+B)⋅x2+(2A+B+C)⋅x+A≡1
We compare the coefficients (to zero) to get 3 simultaneous linear equations,
A+B=0
2A+B+C=0
A=1
The system of equations is solved easily to yield
A=1
B=−1
C=−1
Therefore,
1x(x+1)2≡1x−1x+1−1(x+1)2 .
Now, to integrate.
∫1x(x+1)2dx=∫(1x−1x+1−1(x+1)2)dx
