How do you integrate #int 1/(s+1)^2# using partial fractions?

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Answer 1 · 2017-10-26T17:03:57

The answer is #=-1/(1+s)+C#

You don't need partial fractions

Perform the substitution

#u=1+s#, #du=ds#

#int(ds)/(1+s)^2=int(u)^-2du#

#=-1/(u)#

#=-1/(1+s)+C#

Answer 2 · 2017-10-26T17:18:19

Partial fraction decomposition does not help; it gives you: #1/(s+1)^2#

The integral #int 1/(s+1)^2ds# is best integrated by "u" substitution.

Set up the expansion equation:

#1/(s+1)^2 = A/(s+1)+B/(s+1)^2#

Multiply both sides by #(s+1)^2#:

#1 = A(s+1)+B#

#A = 0, B=1#

The decomposition is the same as the original:

#1/(s+1)^2 = 1/(s+1)^2#

Returning to the integral:

#int 1/(s+1)^2 ds#

let #u = s+1#, then #du = ds# and the integral becomes:

#int u^-2 du = -u^-1 + C#

Reverse the substitution:

#int 1/(s+1)^2 ds = -(s+1)^-1 + C#