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

2 Answers
George C.
Nov 3, 2016

int color(white)(.)x^2/(x+1)^3color(white)(.) dx = ln abs(x+1)+2/(x+1)-1/(2(x+1)^2)+C

Explanation:

int color(white)(.)x^2/(x+1)^3color(white)(.) dx = int color(white)(.)((x^2+2x+1)-(2x+2)+1)/(x+1)^3color(white)(.) dx

color(white)(int color(white)(.)x^2/(x+1)^3color(white)(.) dx) = int color(white)(.)((x+1)^2-2(x+1)+1)/(x+1)^3color(white)(.) dx

color(white)(int color(white)(.)x^2/(x+1)^3color(white)(.) dx) = int color(white)(.)1/(x+1)-2/(x+1)^2+1/(x+1)^3color(white)(.) dx

color(white)(int color(white)(.)x^2/(x+1)^3color(white)(.) dx) = ln abs(x+1)+2/(x+1)-1/(2(x+1)^2)+C

Narad T.
Nov 3, 2016

The answer is =-1/(2(x+1)^2)+(2)/(x+1)+ln(x+1)+C

Explanation:

Let's do the decomposition in partial fractions
x^2/(x+1)^3=A/(x+1)^3+B/(x+1)^2+C/(x+1)
=(A+B(x+1)+C(x+1)^2)/(x+1)^3
We need to find A,B,C
So x^2=A+B(x+1)+C(x+1)^2
coefficients of x^2=>C=1
let x=-1=>A=1
Coefficients of x, 0=B+2C=>B=-2
:.x^2/(x+1)^3=1/(x+1)^3-2/(x+1)^2+1/(x+1)
So int(x^2dx)/(x+1)^3=intdx/(x+1)^3-int(2dx)/(x+1)^2+intdx/(x+1)
We use intx^ndx=x^(n+1)/(n+1) and intdx/x=lnx
int(x^2dx)/(x+1)^3=-1/(2(x+1)^2)+(2)/(x+1)+ln(x+1)+C

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