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

The integral of (x+1)/((2x+1)(x-1)) is -1/6 ln|2x+1|+2/3 ln|x-1|+C

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

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

- \frac{1}{6} ln ∣ 2 x + 1 ∣ + \frac{2}{3} ln ∣ x - 1 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{u}{( 2 u - 1 ) ( u - 2 )} d u

= \int \frac{u}{( 2 u - 1 ) ( u - 2 )} d u

Take the partial fraction of

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

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

\int \frac{1}{3 ( 2 u - 1 )} d u = \frac{1}{6} ln ∣ 2 u - 1 ∣

\int \frac{2}{3 ( u - 2 )} d u = \frac{2}{3} ln ∣ u - 2 ∣

= - \frac{1}{6} ln ∣ 2 u - 1 ∣ + \frac{2}{3} ln ∣ u - 2 ∣

Substitute back

u = x + 1

= - \frac{1}{6} ln ∣ 2 (x + 1) - 1 ∣ + \frac{2}{3} ln ∣ x + 1 - 2 ∣

Simplify

- \frac{1}{6} ln ∣ 2 (x + 1) - 1 ∣ + \frac{2}{3} ln ∣ x + 1 - 2 ∣ : - \frac{1}{6} ln ∣ 2 x + 1 ∣ + \frac{2}{3} ln ∣ x - 1 ∣

= - \frac{1}{6} ln ∣ 2 x + 1 ∣ + \frac{2}{3} ln ∣ x - 1 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ 2 x + 1 ∣ + \frac{2}{3} ln ∣ x - 1 ∣ + C

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What is the integral of (x+1)/((2x+1)(x-1)) ?
The integral of (x+1)/((2x+1)(x-1)) is -1/6 ln|2x+1|+2/3 ln|x-1|+C