Answer
$\frac{1}{6}\ln \left| {3{t^2} + 6t - 5} \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{t + 1}}{{3{t^2} + 6t - 5}}} dt \cr
& {\text{Let }}u = 3{t^2} + 6t - 5,{\text{ then }}du = \left( {6t + 6} \right)dt \cr
& du = 6\left( {t + 1} \right)dt \cr
& \frac{1}{6}du = \left( {t + 1} \right)dt \cr
& {\text{Applying the substitution}}{\text{, we obtain}} \cr
& \int {\frac{{t + 1}}{{3{t^2} + 6t - 5}}} dt = \int {\frac{1}{u}\left( {\frac{1}{6}} \right)} du \cr
& = \frac{1}{6}\int {\frac{1}{u}} du \cr
& {\text{Integrate}} \cr
& = \frac{1}{6}\ln \left| u \right| + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}u = 3{t^2} + 6t - 5 \cr
& = \frac{1}{6}\ln \left| {3{t^2} + 6t - 5} \right| + C \cr} $$
