How do you find #int (x^2+1)/(x+1)^3dx# using partial fractions?

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Answer 1 · 2018-05-31T17:16:10

#x-2/(x+1)-2ln|x+1|+C#

Note that

#(x^2+1)/(x+1)^2=1-1/(x+1)+2/(x+1)^2#

Answer 2 · 2018-05-31T19:08:55

#ln(x+1)+2/(x+1)-1/(x+1)^2+C#

Due to #x^2+1=x^2+2x+1-2x-2+2# or #(x+1)^2-2*(x+1)+2#

#int (x^2+1)/(x+1)^3*dx#

=#int dx/(x+1)-2int dx/(x+1)^2+2int dx/(x+1)^2#

=#ln(x+1)+2/(x+1)-1/(x+1)^2+C#