How do you find the definite integral of #int 1/(x+2)# from [-1,1]?

1 Answer
Mia
Dec 4, 2016

#ln3#

Explanation:

Let #u=x+2#
#" "#
then #" "du=dx#
#" "#
#int1/(x+2) dx#
#" "#
#=int (du)/u#
#" "#
#=lnu + C#
#" "#
#=lnabs(x+2) + C#
#" "#
#" "#
#int_-1^1 1/(x+2) dx#
#" "#
#=lnabs(1+2) - lnabs(-1+2)#
#" "#
#=lnabs3-lnabs1#
#" "#
#=ln3-ln1#
#" "#
#=ln3-0#
#" "#
#=ln3#

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