Answer
$\int_1^2\frac{dt}{8-3t}=\frac{1}{3}[ln(\frac{5}{2})]$
Work Step by Step
Evaluate the integral $\int_1^2\frac{dt}{8-3t}$.
$\int_1^2\frac{dt}{8-3t}=ln[\frac{8-3t}{-3}]_1^2$
$=-\frac{1}{3}[ln2-ln5]$
Thus, $=\frac{1}{3}[ln5-ln2]$
Hence, $\int_1^2\frac{dt}{8-3t}=\frac{1}{3}[ln(\frac{5}{2})]$
