What is the integral of (14x-1)/((4x+1)(x-2)) ?

The integral of (14x-1)/((4x+1)(x-2)) is 1/2 ln|4x+1|+3ln|x-2|+C

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integral of (14x-1)/((4x+1)(x-2))

\int \frac{14 x - 1}{( 4 x + 1 ) ( x - 2 )} d x

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

Solution steps

\int \frac{14 x - 1}{( 4 x + 1 ) ( x - 2 )} d x

Take the partial fraction of

\frac{14 x - 1}{( 4 x + 1 ) ( x - 2 )} : \frac{2}{4 x + 1} + \frac{3}{x - 2}

= \int \frac{2}{4 x + 1} + \frac{3}{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{2}{4 x + 1} d x + \int \frac{3}{x - 2} d x

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

\int \frac{3}{x - 2} d x = 3 ln ∣ x - 2 ∣

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

Add a constant to the solution

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

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What is the integral of (14x-1)/((4x+1)(x-2)) ?
The integral of (14x-1)/((4x+1)(x-2)) is 1/2 ln|4x+1|+3ln|x-2|+C