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

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Answer 1 · 2016-03-21T07:28:49

We need to find constants A,B,C such as

$(1-x)/(x(x+1)^2)=A/x+B/(x+1)+C/(x+1)^2$

Giving $x$ various values we get that

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

Hence

$int (1-x)/(x(x+1)^2)dx=int 1/xdx-int 1/(x+1)dx-int 2/(x+1)^2dx= lnabsx-lnabs(x+1)+2/(x+1)+c$

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