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

2 Answers
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#

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#