How do you integrate #int (x^3)/(x^2 + 8x + 16)# using partial fractions?

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Answer 1 · 2017-02-01T23:17:59

#1/2x^2-8x+48ln|x+4|+64/(x+4)+c#

The denominator is a perfect square, so you don't use partial fractions. Instead, you substitute #u=x+4#. The integral becomes:
#int(u-4)^3/u^2 du#

#=int(u^3-12u^2+48u-64)/u^2du#

#=intu-12+48/u-64/u^2du#

#=1/2u^2-12u+48ln|u|+64/u+k#

#=1/2(x+4)^2-12(x+4)+48ln|x+4|+64/(x+4)+k#

#=1/2x^2-8x+48ln|x+4|+64/(x+4)+c#

(where #c=8-48+k#)