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

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

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

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

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

Solution steps

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

Expand

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

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

\int \frac{3 x}{( 2 x + 1 ) ( x + 3 )} d x = 3 (\frac{3}{5} ln ∣ x + 3 ∣ - \frac{1}{10} ln ∣ 2 x + 1 ∣ - \frac{1}{2})

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

= 3 (\frac{3}{5} ln ∣ x + 3 ∣ - \frac{1}{10} ln ∣ 2 x + 1 ∣ - \frac{1}{2}) - (\frac{1}{5} ln ∣ 2 x + 1 ∣ - \frac{1}{5} ln ∣ x + 3 ∣)

Simplify

3 (\frac{3}{5} ln ∣ x + 3 ∣ - \frac{1}{10} ln ∣ 2 x + 1 ∣ - \frac{1}{2}) - (\frac{1}{5} ln ∣ 2 x + 1 ∣ - \frac{1}{5} ln ∣ x + 3 ∣) : 2 ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ 2 x + 1 ∣ - \frac{3}{2}

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

Add a constant to the solution

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

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