$\int \frac{2 x + 3}{3 x + 2} d x$ is equal to

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Solution

Given the integral,
$\int \frac{2 x + 3}{3 x + 2} d x$
Let,
$u = 3 x + 2 \Rightarrow \frac{d u}{d x} = 3 \Rightarrow d x = \frac{1}{3} d u$
Substituting these values in the given integral we get,
$\int \frac{2 x + 3}{3 x + 2} d x = \frac{1}{9} \int \frac{2 u + 5}{u} d u$
For,
$\int \frac{2 u + 5}{u} d u = \int (\frac{5}{u} + 2) d u = 5 \int \frac{1}{u} d u + 2 \int (1) d u = 5 ln (u) + 2 u ∴ \frac{1}{9} \int \frac{2 u + 5}{u} d u = \frac{5 ln (u)}{9} + \frac{2 u}{9} = \frac{5 ln (3 x + 2)}{9} + \frac{2 (3 x + 2)}{9}$
Hence, $\int \frac{2 x + 3}{3 x + 2} d x = \frac{5 ln (3 x + 2)}{9} + \frac{2 (3 x + 2)}{9} + C .$

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