What is the integral of (8x-10)/((2x-1)(x+1)) ?

The integral of (8x-10)/((2x-1)(x+1)) is -2ln|2x-1|+6ln|x+1|+C

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integral of (8x-10)/((2x-1)(x+1))

\int \frac{8 x - 10}{( 2 x - 1 ) ( x + 1 )} d x

- 2 ln ∣ 2 x - 1 ∣ + 6 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{8 x - 10}{( 2 x - 1 ) ( x + 1 )} d x

Take the partial fraction of

\frac{8 x - 10}{( 2 x - 1 ) ( x + 1 )} : - \frac{4}{2 x - 1} + \frac{6}{x + 1}

= \int - \frac{4}{2 x - 1} + \frac{6}{x + 1} 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{4}{2 x - 1} d x + \int \frac{6}{x + 1} d x

\int \frac{4}{2 x - 1} d x = 2 ln ∣ 2 x - 1 ∣

\int \frac{6}{x + 1} d x = 6 ln ∣ x + 1 ∣

= - 2 ln ∣ 2 x - 1 ∣ + 6 ln ∣ x + 1 ∣

Add a constant to the solution

= - 2 ln ∣ 2 x - 1 ∣ + 6 ln ∣ x + 1 ∣ + C

What is the integral of (8x-10)/((2x-1)(x+1)) ?
The integral of (8x-10)/((2x-1)(x+1)) is -2ln|2x-1|+6ln|x+1|+C