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