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

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Answer 1 · 2016-06-16T00:59:10

You don't use partial fraction, do substitution

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

#int 1/u^2 du#

which is

#[-1/u]#

and subsitute back you have :

#[-1/(s+1)]#