Answer
$$\frac{1}{5}\ln \left| {\frac{{\tan \left( {x/2} \right) - 1/3}}{{\tan \left( {x/2} \right) + 3}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{4\sin x - 3\cos x}}} \cr
& {\text{Use the substitutions }}\left( 5 \right){\text{ from the page 528}}: \cr
& \,\,u = \tan \left( {\frac{x}{2}} \right){\text{,}}\,\,\,\,dx = \frac{2}{{1 + {u^2}}}du,\,\,\,\sin x = \frac{{2u}}{{1 + {u^2}}}\,\,\,{\text{and}}\,\,\cos x = \frac{{1 - {u^2}}}{{1 + {u^2}}} \cr
& \cr
& \int {\frac{{dx}}{{4\sin x - 3\cos x}}} = \int {\frac{{\frac{2}{{1 + {u^2}}}du}}{{4\left( {\frac{{2u}}{{1 + {u^2}}}} \right) - 3\left( {\frac{{1 - {u^2}}}{{1 + {u^2}}}} \right)}}} \cr
& {\text{simplify}} \cr
& = \int {\frac{{\frac{2}{{1 + {u^2}}}du}}{{\frac{{8u}}{{1 + {u^2}}} - \frac{{3\left( {1 - {u^2}} \right)}}{{1 + {u^2}}}}}} \cr
& = \int {\frac{2}{{8u - 3 + 3{u^2}}}} du \cr
& = \int {\frac{2}{{3{u^2} + 8u - 3}}} du \cr
& \cr
& {\text{Complete the square for }}3{u^2} + 8u - 3 \cr
& 3{u^2} + 8u - 3 = 3\left( {{u^2} + \frac{8}{3}u + \frac{{16}}{9}} \right) - 3 - 3\left( {\frac{{16}}{9}} \right) \cr
& 3{u^2} + 8u - 3 = 3{\left( {u + \frac{4}{3}} \right)^2} - \frac{{25}}{2} \cr
& \int {\frac{2}{{3{u^2} + 8u - 3}}} du = \frac{2}{3}\int {\frac{1}{{{{\left( {u + 4/3} \right)}^2} - {{\left( {25/3} \right)}^2}}}} du \cr
& {\text{Let }}t = u + 4/3 \cr
& = \frac{2}{3}\int {\frac{1}{{{t^2} - {{\left( {5/3} \right)}^2}}}} du \cr
& \frac{2}{3}\frac{3}{{2\left( 5 \right)}}\ln \left| {\frac{{u - 5/3}}{{u + 5/3}}} \right| + C \cr
& \frac{1}{5}\ln \left| {\frac{{u + 4/3 - 5/3}}{{u + 4/3 + 5/3}}} \right| + C \cr
& \frac{1}{5}\ln \left| {\frac{{u - 1/3}}{{u + 3}}} \right| + C \cr
& {\text{Where }}u = \tan \left( {\frac{x}{2}} \right) \cr
& \frac{1}{5}\ln \left| {\frac{{\tan \left( {x/2} \right) - 1/3}}{{\tan \left( {x/2} \right) + 3}}} \right| + C \cr} $$
