MHB Sava's question via email about integration with partial fractions.

[Prove It]

Evaluate $\displaystyle \begin{align*} \int{\frac{x^2}{\left( x - 1 \right) ^2 \, \left( x - 2 \right) } \,\mathrm{d}x} \end{align*}$

As there is a repeated root, the partial fraction decomposition we should use is:

$\displaystyle \begin{align*} \frac{A}{x - 1} + \frac{B}{\left( x - 1 \right) ^2 } + \frac{C}{x - 2} &\equiv \frac{x^2}{\left( x - 1 \right) ^2\,\left( x - 2 \right) } \\ \frac{A\,\left( x - 1 \right) \left( x - 2 \right) + B\,\left( x - 2 \right) + C\,\left( x - 1 \right) ^2 }{\left( x -1 \right) ^2 \,\left( x - 2 \right) } &\equiv \frac{x^2}{\left( x - 1 \right) ^2\,\left( x - 2 \right) } \\ A\,\left( x - 1 \right) \left( x - 2 \right) + B\,\left( x - 2 \right) + C\,\left( x - 1 \right) ^2 &\equiv x^2 \end{align*}$

Let $\displaystyle \begin{align*} x = 1 \end{align*}$ to find $\displaystyle \begin{align*} -B = 1 \implies B = -1 \end{align*}$

Let $\displaystyle \begin{align*} x = 2 \end{align*}$ to find $\displaystyle \begin{align*} C = 4 \end{align*}$

Substitute B and C back in:

$\displaystyle \begin{align*} A\,\left( x - 1 \right) \left( x - 2 \right) - \left( x - 2 \right) + 4\,\left( x - 1 \right) ^2 &\equiv x^2 \end{align*}$

Let $\displaystyle \begin{align*} x = 0 \end{align*}$ to find $\displaystyle \begin{align*} 2\,A + 2 + 4 = 0 \implies 2\,A = -6 \implies A = -3 \end{align*}$.

So that means

$\displaystyle \begin{align*} \int{ \frac{x^2}{\left( x - 1 \right) ^2\,\left( x - 2 \right) } \,\mathrm{d}x} &= \int{ \left[ -\frac{3}{x - 1} - \frac{1}{\left( x - 1 \right) ^2 } + \frac{4}{x - 2} \right] \,\mathrm{d}x } \\ &= -3\ln{\left| x - 1 \right| } + \frac{1}{x - 1} + 4\ln{\left| x - 2 \right| } + C \end{align*}$

 

[chwala]

As there is a repeated root, the partial fraction decomposition we should use is:

$\displaystyle \begin{align*} \frac{A}{x - 1} + \frac{B}{\left( x - 1 \right) ^2 } + \frac{C}{x - 2} &\equiv \frac{x^2}{\left( x - 1 \right) ^2\,\left( x - 2 \right) } \\ \frac{A\,\left( x - 1 \right) \left( x - 2 \right) + B\,\left( x - 2 \right) + C\,\left( x - 1 \right) ^2 }{\left( x -1 \right) ^2 \,\left( x - 2 \right) } &\equiv \frac{x^2}{\left( x - 1 \right) ^2\,\left( x - 2 \right) } \\ A\,\left( x - 1 \right) \left( x - 2 \right) + B\,\left( x - 2 \right) + C\,\left( x - 1 \right) ^2 &\equiv x^2 \end{align*}$

Let $\displaystyle \begin{align*} x = 1 \end{align*}$ to find $\displaystyle \begin{align*} -B = 1 \implies B = -1 \end{align*}$

Let $\displaystyle \begin{align*} x = 2 \end{align*}$ to find $\displaystyle \begin{align*} C = 4 \end{align*}$

Substitute B and C back in:

$\displaystyle \begin{align*} A\,\left( x - 1 \right) \left( x - 2 \right) - \left( x - 2 \right) + 4\,\left( x - 1 \right) ^2 &\equiv x^2 \end{align*}$

Let $\displaystyle \begin{align*} x = 0 \end{align*}$ to find $\displaystyle \begin{align*} 2\,A + 2 + 4 = 0 \implies 2\,A = -6 \implies A = -3 \end{align*}$.

So that means

$\displaystyle \begin{align*} \int{ \frac{x^2}{\left( x - 1 \right) ^2\,\left( x - 2 \right) } \,\mathrm{d}x} &= \int{ \left[ -\frac{3}{x - 1} - \frac{1}{\left( x - 1 \right) ^2 } + \frac{4}{x - 2} \right] \,\mathrm{d}x } \\ &= -3\ln{\left| x - 1 \right| } + \frac{1}{x - 1} + 4\ln{\left| x - 2 \right| } + C \end{align*}$

This is correct!