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

The integral of 1/((2x+1)(x+2)) is 1/3 ln|2x+1|-1/3 ln|2x+4|+C

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

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

\frac{1}{3} ln ∣ 2 x + 1 ∣ - \frac{1}{3} ln ∣ 2 x + 4 ∣ + C

Solution steps

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

Apply u-substitution

= \int \frac{1}{u ( u + 3 )} d u

Take the partial fraction of

\frac{1}{u ( u + 3 )} : \frac{1}{3 u} - \frac{1}{3 ( u + 3 )}

= \int \frac{1}{3 u} - \frac{1}{3 ( u + 3 )} d u

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

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

\int \frac{1}{3 ( u + 3 )} d u = \frac{1}{3} ln ∣ u + 3 ∣

= \frac{1}{3} ln ∣ u ∣ - \frac{1}{3} ln ∣ u + 3 ∣

Substitute back

u = 2 x + 1

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

Simplify

\frac{1}{3} ln ∣ 2 x + 1 ∣ - \frac{1}{3} ln ∣ 2 x + 1 + 3 ∣ : \frac{1}{3} ln ∣ 2 x + 1 ∣ - \frac{1}{3} ln ∣ 2 x + 4 ∣

= \frac{1}{3} ln ∣ 2 x + 1 ∣ - \frac{1}{3} ln ∣ 2 x + 4 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 2 x + 1 ∣ - \frac{1}{3} ln ∣ 2 x + 4 ∣ + C

What is the integral of 1/((2x+1)(x+2)) ?
The integral of 1/((2x+1)(x+2)) is 1/3 ln|2x+1|-1/3 ln|2x+4|+C