What is the integral of (3x+2)/((x+1)(2x+3)^2) ?

The integral of (3x+2)/((x+1)(2x+3)^2) is -ln|x+1|+ln|2x+3|-5/(2(2x+3))+C

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integral of (3x+2)/((x+1)(2x+3)^2)

\int \frac{3 x + 2}{( x + 1 ) ( 2 x + 3 ) ^{2}} d x

- ln ∣ x + 1 ∣ + ln ∣ 2 x + 3 ∣ - \frac{5}{2 ( 2 x + 3 )} + C

Solution steps

\int \frac{3 x + 2}{( x + 1 ) ( 2 x + 3 ) ^{2}} d x

Take the partial fraction of

\frac{3 x + 2}{( x + 1 ) ( 2 x + 3 ) ^{2}} : - \frac{1}{x + 1} + \frac{2}{2 x + 3} + \frac{5}{( 2 x + 3 ) ^{2}}

= \int - \frac{1}{x + 1} + \frac{2}{2 x + 3} + \frac{5}{( 2 x + 3 ) ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{x + 1} d x + \int \frac{2}{2 x + 3} d x + \int \frac{5}{( 2 x + 3 ) ^{2}} d x

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

\int \frac{2}{2 x + 3} d x = ln ∣ 2 x + 3 ∣

\int \frac{5}{( 2 x + 3 ) ^{2}} d x = - \frac{5}{2 ( 2 x + 3 )}

= - ln ∣ x + 1 ∣ + ln ∣ 2 x + 3 ∣ - \frac{5}{2 ( 2 x + 3 )}

Add a constant to the solution

= - ln ∣ x + 1 ∣ + ln ∣ 2 x + 3 ∣ - \frac{5}{2 ( 2 x + 3 )} + C

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What is the integral of (3x+2)/((x+1)(2x+3)^2) ?
The integral of (3x+2)/((x+1)(2x+3)^2) is -ln|x+1|+ln|2x+3|-5/(2(2x+3))+C