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

1 Answer

Nov 30, 2016

#int_(-2)^2 frac 1 ((x+1)^3) dx =-4/9#

Explanation:

#int_(-2)^2 frac 1 ((x+1)^3) dx = int_(-2)^2 frac 1 ((x+1)^3) d(x+1) =#

#= -1/2 frac 1 ((1+x)^2) |_(x=-2)^(x=2) = #

#= -1/2 frac 1 ((-2+1)^2) +1/2 frac 1 ((2+1)^2) = -1/2+1/2 * 1/9 = #

# =1/2(1/9 -1) = -1/2 * 8/9 =-4/9#

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