How do you integrate #int (x^3-x^2)/ (x+3)^4# using partial fractions?

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How do you integrate #int (x^3-x^2)/ (x+3)^4# using partial fractions?

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Answer 1 · 2017-01-18T12:24:04

#ln|x+3| +10/(x+3)-33/(2(x+3)^2)+12/(x+3)^3+c#

Explanation:

Let #u=x+3#, #(du)/(dx)=1#.
#int((u-3)^3-(u-3)^2)/u^4 du#
#=int(u^3-10u^2+33u-36)/u^4du#
#=int u^-1-10u^-2+33u^-3-36u^-4dx#
#=ln|u|+10/u-33/(2u^2)+12/u^3+c#

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How do you integrate #int (x^3-x^2)/ (x+3)^4# using partial fractions?

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