Integrating a Rational Function Whose Denominator Contains a Repeated Linear Factor

Find \(\int \dfrac{dx}{x(x-1)^{2}}\).

Solution  Since \(x\) is a distinct linear factor of the denominator \(q\), and \((x-1)^{2}\) is a repeated linear factor of the denominator, the decomposition of \(\dfrac{1}{x(x-1) ^{2}}\) into partial fractions has the three terms \(\dfrac{A}{x}\), \(\dfrac{B}{x-1},\) and \(\dfrac{C}{(x-1) ^{2}}\). \[ \begin{eqnarray*} \begin{array}{rl@{\!\!\!\qquad}l} \dfrac{1}{x(x-1)^{2}} &= \dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{(x-1)^{2}} \quad &{\color{#0066A7}{\hbox{Write the identity.}}} \\ 1 &= A(x-1)^{2}+B\cdot x(x-1)+C\cdot x\quad &{\color{#0066A7}{\hbox{Multiply both sides by }{x} (x-1)^{2}.}} \end{array} \end{eqnarray*} \]

We find \(A,\) \(B,\) and \(C\) by choosing values of \(x\) that cause one or more terms to drop out. When \(x=1,\) we have \(1=C\cdot 1,\) so \(C=1\). When \(x=0,\) we have \(1=A(0-1)^{2},\) so \(A=1.\) Now using \(A=1\) and \(C=1,\) we have \[ \begin{equation*} 1=(x-1)^{2}+B\cdot x(x-1)+ 1 \cdot x \end{equation*} \]

Suppose we let \(x=2.\) (Any choice other than \(0\) and \(1\) will also work.) Then \[ \begin{eqnarray*} 1 &=&1+2B+2 \\ B &=&-1 \end{eqnarray*} \]

Then \[ \begin{array}{l@{\qquad}l} \dfrac{1}{x(x-1)^{2}}=\dfrac{1}{x}+\dfrac{-1}{(x-1) }+\dfrac{1}{(x-1)^{2}} \quad{\color{#0066A7}{\hbox{\(A=1\)}}\quad \quad{\color{#0066A7}{\hbox{\(B=-1\)}}}\quad \quad{\color{#0066A7}{\hbox{\(C=1\)}}}} \end{array} \]

503

So, \[ \begin{eqnarray*} \int \dfrac{dx}{x(x-1)^{2}} &=& \int \dfrac{dx}{x}- \int \dfrac{dx}{x-1}+ \int \dfrac{dx}{(x-1)^{2}}\\[7pt] &=& \ln \vert x \vert -\ln \vert x-1\vert -\dfrac{1}{x-1} +C_{1} \end{eqnarray*} \]