## Calculus: Early Transcendentals (2nd Edition)

$$\frac{1}{6}\ln \left( {\frac{1}{2}} \right)$$
\eqalign{ & \int_0^1 {\frac{{dt}}{{{t^2} - 9}}} \cr & {\text{partial fraction decomposition}} \cr & \frac{1}{{{t^2} - 9}} = \frac{1}{{\left( {t + 3} \right)\left( {t - 3} \right)}} \cr & \frac{1}{{\left( {t - 3} \right)\left( {t + 3} \right)}} = \frac{A}{{t - 3}} + \frac{B}{{t + 3}} \cr & 1 = A\left( {t + 3} \right) + B\left( {t - 3} \right) \cr & {\text{for }}t = 3 \cr & 1 = A\left( {3 + 3} \right) + B\left( 0 \right) \cr & 1/6 = A \cr & {\text{for }}t = - 3 \cr & 1 = A\left( 0 \right) + B\left( { - 3 - 3} \right) \cr & 1 = - 6B \cr & - 1/6 = B \cr & {\text{substituting the values}} \cr & \frac{A}{{t - 3}} + \frac{B}{{t + 3}} = \frac{{1/6}}{{t - 3}} - \frac{{1/6}}{{t + 3}} \cr & \int_0^1 {\frac{{dt}}{{{t^2} - 9}}} = \int_0^1 {\left( {\frac{{1/6}}{{t - 3}} - \frac{{1/6}}{{t + 3}}} \right)} dt \cr & = \frac{1}{6}\int_0^1 {\left( {\frac{1}{{t - 3}} - \frac{1}{{t + 3}}} \right)} dt \cr & {\text{integrating}} \cr & = \frac{1}{6}\left[ {\ln \left| {t - 3} \right| - \ln \left| {t + 3} \right|} \right]_0^1 \cr & {\text{evaluating limits}} \cr & = \frac{1}{6}\left[ {\ln \left| {1 - 3} \right| - \ln \left| {1 + 3} \right|} \right] - \frac{1}{6}\left[ {\ln \left| {0 - 3} \right| - \ln \left| {0 + 3} \right|} \right] \cr & = \frac{1}{6}\left[ {\ln 2 - \ln 4} \right] - \frac{1}{6}\left[ {\ln 3 - \ln 3} \right] \cr & = \frac{1}{6}\ln \left( {\frac{2}{4}} \right) \cr & = \frac{1}{6}\ln \left( {\frac{1}{2}} \right) \cr}