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 $(x+1)^2-2*(x+1)+2$

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

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

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

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