What is the integral of (8x+6)/(2x^2+3x-2) ?

The integral of (8x+6)/(2x^2+3x-2) is 2ln|2x-1|+2ln|x+2|+C

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integral of (8x+6)/(2x^2+3x-2)

\int \frac{8 x + 6}{2 x ^{2} + 3 x - 2} d x

2 ln ∣ 2 x - 1 ∣ + 2 ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{8 x + 6}{2 x ^{2} + 3 x - 2} d x

Take the partial fraction of

\frac{8 x + 6}{2 x ^{2} + 3 x - 2} : \frac{4}{2 x - 1} + \frac{2}{x + 2}

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

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

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

= 2 ln ∣ 2 x - 1 ∣ + 2 ln ∣ x + 2 ∣

Add a constant to the solution

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

What is the integral of (8x+6)/(2x^2+3x-2) ?
The integral of (8x+6)/(2x^2+3x-2) is 2ln|2x-1|+2ln|x+2|+C