How do you integrate #int (x^2 + 5x - 7) /( x^2 (x+ 1)^2)# using partial fractions?

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Answer 1 · 2017-02-03T11:03:53

The answer is #=7/x+19ln(|x|)+11/(x+1)-19ln(|x+1|)+C#

Let's perform the decomposition into partial fractions

#(x^2+5x-7)/(x^2(x+1)^2)=A/x^2+B/x+C/(x+1)^2+D/(x+1)#

#=(A(x+1)^2+Bx(x+1)^2+Cx^2+Dx^2(x+1))/((x^2(x+1)^2))#

As the denominators are the same, we can compare the numerators

#x^2+5x-7=A(x+1)^2+Bx(x+1)^2+Cx^2+Dx^2(x+1)#

Let #x=0#, #=>#, #-7=A#

Let #x=-1#, #=>#, #-11=C#

Coefficients of #x^3#, #=>#, #0=B+D#

Coefficients of #x#, #=>#, #5=2A+B#

#B=5-2A=5+14=19#

#D=-B=-19#

Therefore,

#(x^2+5x-7)/(x^2(x+1)^2)=-7/x^2+19/x-11/(x+1)^2-19/(x+1)#

So,

#int((x^2+5x-dx)/(x^2(x+1)^2)=-7intdx/x^2+19intdx/x-11intdx/(x+1)^2-19intdx/(x+1)#

#=7/x+19ln(|x|)+11/(x+1)-19ln(|x+1|)+C#