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

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

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

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

$= 0$

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