$\int {\frac{{2x + 3}}{{{x^2} + 3x}}} dx = \log \left| {{x^2} + 3x} \right| + C$
$\int {\frac{{2x + 3}}{{{x^2} + 3x}}} dx = \log \left| {{x^2} + 3x} \right| + C$
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Let $I = \int {\frac{{2x + 3}}{{{x^2} + 3x}}} dx$
Let's put ${x^2} + 3x = t$
$\Rightarrow$ $(2x + 3)dx = dt$
therefore,$I = \int {\frac{1}{t}} dt = \log |t| + C$
$= \log \left| {\left( {{x^2} + 3x} \right)} \right| + C$
Evaluate the following :
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