What is #int (4+u)/u^3 du#?

1 Answer
Alan N.
Mar 7, 2018

#I= -{(u+2)/u^2}+C#

Explanation:

Let #I= int(4+u)/u^3du#

#I= int(4/u^3 + 1/u^2)du#

Apply linearity.

#I= int(4/u^3)du + int(1/u^2)du#

#= 4intu^-3 du + intu^-2 du#

Apply power rule.

#I= 4xxu^-2/(-2) + u^-1/-1 +C#

#= -2/u^2 - 1/u+C#

#-{(u+2)/u^2}+C#

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