$\int {\frac{{2x - 1}}{{2x + 3}}} dx = x - \log \left| {{{(2x + 3)}^2}} \right| + C$
$\int {\frac{{2x - 1}}{{2x + 3}}} dx = x - \log \left| {{{(2x + 3)}^2}} \right| + C$
Official Solution
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Let $I = \int {\frac{{2x - 1}}{{2x + 3}}} dx = \int {\frac{{2x + 3 - 3 - 1}}{{2x + 3}}} dx$
$= \int 1 dx - 4\int {\frac{1}{{2x + 3}}} dx = x - \int {\frac{4}{{2\left( {x + \frac{3}{2}} \right)}}} dx$
$= x - 2\log + \left| {\left( {x + \frac{3}{2}} \right)} \right|{C^\prime } = x - 2\log \left| {\left( {\frac{{2x + 3}}{2}} \right)} \right| + {C^\prime }$
$= x - 2\log |(2x + 3)| + 2\log 2 + {C^\prime }$
$= x - \log \left| {{{(2x + 3)}^2}} \right| + C$
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