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

##### 1 Answer

I got

With **partial fractions** where the denominator has a *multiplicity* of

#int (x^2)/(x-1)^3dx = int A/(x-1) + B/(x-1)^2 + C/(x-1)^3dx#

Now, we should get **common denominators** so that we can set the resultant fraction equal to the starting integrand.

That way, we can equate each *coefficient* to the *numerator coefficients* and find

Ignoring the integral symbols for now, we can focus on the integrand:

#= (A(x-1)^2)/(x-1)^3 + (B(x-1))/(x-1)^3 + C/(x-1)^3#

Combine the fractions:

#= (A(x-1)^2 + B(x-1) + C)/(x-1)^3#

Distribute the numerator terms:

#= (Ax^2 - 2Ax + A + Bx - B + C)/cancel((x-1)^3) = (x^2)/cancel((x-1)^3)#

Next, we can rearrange the terms to turn this into the form

Remember that this form **must** include the

#(A)x^2 \mathbf(color(highlight)(+)) (-2A + B)x \mathbf(color(highlight)(+)) (A - B + C) = x^2#

Thus, we have the following system of equations:

#color(green)(A = 1)#

#-2A + B = 0#

#A - B + C = 0#

Knowing

#2A = B => color(green)(B = 2)#

#A - B + C = 0 => 1 - 2 + C = 0 => color(green)(C = 1)#

Thus, our resultant integrals are calculated as follows:

#color(blue)(int (x^2)/(x-1)^3dx) = int 1/(x-1) + 2/(x-1)^2 + 1/(x-1)^3dx#

#= int 1/(x-1)dx + 2int1/(x-1)^2dx + int1/(x-1)^3dx#

#= int 1/(x-1)dx + 2int(x-1)^(-2)dx + int(x-1)^(-3)dx#

#= ln|x-1| + (2(x-1)^(-1))/(-1) + ((x-1)^(-2))/(-2)#

#= color(blue)(ln|x-1| - 2/(x-1) - 1/(2(x-1)^2) + C)#