Show that \[\int {\frac{{2x – 1}}{{2x + 3}}} dx = x – \log \left| {{{(2x + 3)}^2}} \right| + C\] ~~~~~[NCERT Exemp. Q. 1,Page 163 ]

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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