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

May 30, 2016

I got $\ln | x - 1 | - \frac{2}{x - 1} - \frac{1}{2 {\left(x - 1\right)}^{2}} + C$.

With partial fractions where the denominator has a multiplicity of $3$ (that is, it is a perfect cube), this factors into:

$\int \frac{{x}^{2}}{x - 1} ^ 3 \mathrm{dx} = \int \frac{A}{x - 1} + \frac{B}{x - 1} ^ 2 + \frac{C}{x - 1} ^ 3 \mathrm{dx}$

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 $A$, $B$, and $C$.

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

$= \frac{A {\left(x - 1\right)}^{2}}{x - 1} ^ 3 + \frac{B \left(x - 1\right)}{x - 1} ^ 3 + \frac{C}{x - 1} ^ 3$

Combine the fractions:

$= \frac{A {\left(x - 1\right)}^{2} + B \left(x - 1\right) + C}{x - 1} ^ 3$

Distribute the numerator terms:

$= \frac{A {x}^{2} - 2 A x + A + B x - B + C}{\cancel{{\left(x - 1\right)}^{3}}} = \frac{{x}^{2}}{\cancel{{\left(x - 1\right)}^{3}}}$

Next, we can rearrange the terms to turn this into the form ${a}_{0} {x}^{2} \textcolor{h i g h l i g h t}{+} {a}_{1} x \textcolor{h i g h l i g h t}{+} {a}_{2}$, which is the standard form of a quadratic equation.

Remember that this form must include the $\textcolor{h i g h l i g h t}{\text{addition}}$ of all ordered terms as a group; no subtractions!

$\left(A\right) {x}^{2} \setminus m a t h b f \left(\textcolor{h i g h l i g h t}{+}\right) \left(- 2 A + B\right) x \setminus m a t h b f \left(\textcolor{h i g h l i g h t}{+}\right) \left(A - B + C\right) = {x}^{2}$

Thus, we have the following system of equations:

$\textcolor{g r e e n}{A = 1}$
$- 2 A + B = 0$
$A - B + C = 0$

Knowing $A$ already, we can easily solve this to get:

$2 A = B \implies \textcolor{g r e e n}{B = 2}$
$A - B + C = 0 \implies 1 - 2 + C = 0 \implies \textcolor{g r e e n}{C = 1}$

Thus, our resultant integrals are calculated as follows:

$\textcolor{b l u e}{\int \frac{{x}^{2}}{x - 1} ^ 3 \mathrm{dx}} = \int \frac{1}{x - 1} + \frac{2}{x - 1} ^ 2 + \frac{1}{x - 1} ^ 3 \mathrm{dx}$

$= \int \frac{1}{x - 1} \mathrm{dx} + 2 \int \frac{1}{x - 1} ^ 2 \mathrm{dx} + \int \frac{1}{x - 1} ^ 3 \mathrm{dx}$

$= \int \frac{1}{x - 1} \mathrm{dx} + 2 \int {\left(x - 1\right)}^{- 2} \mathrm{dx} + \int {\left(x - 1\right)}^{- 3} \mathrm{dx}$

$= \ln | x - 1 | + \frac{2 {\left(x - 1\right)}^{- 1}}{- 1} + \frac{{\left(x - 1\right)}^{- 2}}{- 2}$

$= \textcolor{b l u e}{\ln | x - 1 | - \frac{2}{x - 1} - \frac{1}{2 {\left(x - 1\right)}^{2}} + C}$