Evaluating an Integral!?

Question

Integral((ds)/(s^2(s-1)^2))

Answer 1

∫1/(s²(s-1)²) ds

This is a rational function, already factored, and having only powers of linear terms. In this case, the calculus isn't hard at all. The hard part is obtaining the partial fraction decomposition, which we do as follows:

1/(s²(s-1)²) = A/s + B/s² + C/(s-1) + D/(s-1)²
1 = As(s-1)² + B(s-1)² + Cs²(s-1) + Ds²

Setting s=0 yields:

1 = B(-1)² = B

Setting s=1 yields:

1=D

Setting s=2 yields:

1 = 2A + B + 4C + 4D = 2A + 4C + 5

Partially expanding the sum on the right and equating the coefficients of s³ yields:

0 = A+C
A=-C

Plugging this into the previous equation yields:

1 = -2C + 4C + 5
-4 = 2C
C=-2

Thus we have:

1/(s²(s-1)²) = 2/s + 1/s² - 2/(s-1) + 1/(s-1)²

Plugging this into the integral:

∫2/s + 1/s² - 2/(s-1) + 1/(s-1)² ds
2 ln |s| - 1/s - 2 ln |s-1| - 1/(s-1) + C

And we are done.

Answer 2

Ugly!

-(2x(x-1)*ln(x-1) - 2x(x-1)*ln(x) + 2x - 1)/(x(x-1))

Ugg-lee.