Setting the numerators equal gives,

\[{w^2} + 7w = A\left( {w - 1} \right)\left( {w - 4} \right) + B\left( {w + 2} \right)\left( {w - 4} \right) + C\left( {w + 2} \right)\left( {w - 1} \right)\] Show Step 3

We can use the “trick” discussed in the notes to easily get the coefficients in this case so let’s do that. Here is that work.

\[\begin{align*}w = & \; 1 \,\,\,\,\,\, : & 8 & = - 9B\\ & & & \\ w = & \, 4 \,\,\,\,\,\, : & 44 & = 18C\\ & & & \\ w = & - 2: & - 10 & = 18A\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned}& {A = - \frac{5}{9}}\\ & {B = - \frac{8}{9}}\\ & {C = \frac{{22}}{9}}\end{aligned}\]

The partial fraction form of the integrand is then,

\[\frac{{{w^2} + 7w}}{{\left( {w + 2} \right)\left( {w - 1} \right)\left( {w - 4} \right)}} = \frac{{ - \frac{5}{9}}}{{w + 2}} - \frac{{\frac{8}{9}}}{{w - 1}} + \frac{{\frac{{22}}{9}}}{{w - 4}}\] Show Step 4

We can now do the integral.

\[\begin{align*}\int_{{ - 1}}^{0}{{\frac{{{w^2} + 7w}}{{\left( {w + 2} \right)\left( {w - 1} \right)\left( {w - 4} \right)}}\,dw}} & = \int_{{ - 1}}^{0}{{\frac{{ - \frac{5}{9}}}{{w + 2}} - \frac{{\frac{8}{9}}}{{w - 1}} + \frac{{\frac{{22}}{9}}}{{w - 4}}\,dw}}\\ & = \left. {\left( { - \frac{5}{9}\ln \left| {w + 2} \right| - \frac{8}{9}\ln \left| {w - 1} \right| + \frac{{22}}{9}\ln \left| {w - 4} \right|} \right)} \right|_{ - 1}^0\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{22}}{9}\ln \left( 4 \right) + \frac{3}{9}\ln \left( 2 \right) - \frac{{22}}{9}\ln \left( 5 \right) = \frac{{47}}{9}\ln \left( 2 \right) - \frac{{22}}{9}\ln \left( 5 \right)}}\end{align*}\]

Note that we used a quick logarithm property to combine the first two logarithms into a single logarithm. You should probably review your logarithm properties if you don’t recognize the one that we used. These kinds of property applications can really simplify your work on occasion if you know them!