How do you integrate #int (x^2)/(x+1)^3 dx# using partial fractions?
2 Answers
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#
The answer is
Explanation:
Let's do the decomposition in partial fractions
We need to find A,B,C
So
coefficients of
let
Coefficients of x,
So
We use