How do you evaluate the integral #int 1/(x+1)^(1/3)dx# from -2 to 0?

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Answer 1 · 2016-07-17T17:40:42

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the interval is placed symmetrically about the 2 sided singularity so the areas seem to net off

#int_(-2)^0 1/(x+1)^(1/3)dx#

first note the singularity at x = -1, so through caution i'll do it like this .... probs OTT

#= int_(-2)^(x to -1^-) (x+1)^(-1/3)dx + int_(x to -1^+)^(0) (x+1)^(-1/3)dx#

#=3/2 [ (x+1)^(2/3) ]\_(-2)^(x to -1^-) + 3/2[ (x+1)^(2/3) ]_(x to -1^+)^(0) #

#=3/2 [ 0 - (-1)^(2/3) ] + 3/2[ (1)^(2/3) - 0] #

#= 0#