Answer
$$ - \frac{1}{{\ln 3}}\ln \left| {3 - {3^x}} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{3^x}}}{{3 - {3^x}}}} dx \cr
& {\text{set }}u = 3 - {3^x}{\text{ then }}\frac{{du}}{{dx}} = - {3^x}\ln 3 \to dx = - \frac{1}{{{3^x}\ln 3}}du \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{{3^x}}}{{3 - {3^x}}}} dx = \int {\frac{{{3^x}}}{u}} \left( { - \frac{1}{{{3^x}\ln 3}}du} \right) \cr
& {\text{cancel the common factor}} \cr
& = \int {\frac{1}{u}} \left( { - \frac{1}{{\ln 3}}du} \right) \cr
& = - \frac{1}{{\ln 3}}\int {\frac{1}{u}} du \cr
& {\text{integrating}} \cr
& = - \frac{1}{{\ln 3}}\ln \left| u \right| + C \cr
& {\text{replace }}3 - {3^x}{\text{ for }}u \cr
& = - \frac{1}{{\ln 3}}\ln \left| {3 - {3^x}} \right| + C \cr} $$
