Find $\int \dfrac{x\,dx}{x^{2}-5x+6}$.

Solution The integrand is a proper rational function in lowest terms. We begin by factoring the denominator: $x^{2}-5x+6=(x-2)(x-3)$. Since the factors are linear and distinct, we apply Case 1 and allow for the terms $\dfrac{A}{x-2}$ and $\dfrac{B}{x-3}$. $$\dfrac{x}{(x-2)(x-3)}=\dfrac{A}{x-2}+\dfrac{B}{x-3}$$

Now we clear fractions by multiplying both sides of the equation by $(x-2)(x-3).$ $$\begin{eqnarray*} \begin{array}{ll@{\qquad}l} x &= A(x-3)+B(x-2) \\[3pt] x &= (A+B)x-( 3A+2B) & {\color{#0066A7}{\hbox{Group like terms.}}} \end{array} \end{eqnarray*}$$

This is an identity in $x$, so the coefficients of like powers of $x$ must be equal. $$\begin{eqnarray*} \begin{array}{ll@{\qquad}l} 1 &= A+B & {\color{#0066A7}{\hbox{The coefficient of }{x}\hbox{ equals 1.}}} \\[3pt] 0 &= -3A-2B & {\color{#0066A7}{\hbox{The constant term on the left is 0.}}} \end{array} \end{eqnarray*}$$

This is a system of two equations containing two variables. Solving the second equation for $B,$ we get $B = -\dfrac{3}{2}A.$ Substituting for $B$ in the first equation produces the solution $A$ = -2 from which $B$ = 3. So, $$\dfrac{x}{(x-2)(x-3)}=\dfrac{-2}{x-2}+\dfrac{3}{x-3}$$

501

Then $$\begin{eqnarray*} \int \dfrac{x}{(x-2)(x-3)}dx &=& \int \dfrac{-2}{x-2}\,dx+\int \dfrac{3}{x-3} \,dx \\ &=& -2\,\ln \,\vert x-2 \vert +3\ln \,\vert x-3 \vert +C=\ln \left\vert \dfrac{(x-3)^{3}}{(x-2)^{2}}\right\vert +C \end{eqnarray*}$$