What is the integral of (5x-1)/((2x+1)(x-1)) ?

The integral of (5x-1)/((2x+1)(x-1)) is 7/6 ln|2x+1|+4/3 ln|x-1|+C

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integral of (5x-1)/((2x+1)(x-1))

\int \frac{5 x - 1}{( 2 x + 1 ) ( x - 1 )} d x

\frac{7}{6} ln ∣ 2 x + 1 ∣ + \frac{4}{3} ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{5 x - 1}{( 2 x + 1 ) ( x - 1 )} d x

Take the partial fraction of

\frac{5 x - 1}{( 2 x + 1 ) ( x - 1 )} : \frac{7}{3 ( 2 x + 1 )} + \frac{4}{3 ( x - 1 )}

= \int \frac{7}{3 ( 2 x + 1 )} + \frac{4}{3 ( 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{7}{3 ( 2 x + 1 )} d x + \int \frac{4}{3 ( x - 1 )} d x

\int \frac{7}{3 ( 2 x + 1 )} d x = \frac{7}{6} ln ∣ 2 x + 1 ∣

\int \frac{4}{3 ( x - 1 )} d x = \frac{4}{3} ln ∣ x - 1 ∣

= \frac{7}{6} ln ∣ 2 x + 1 ∣ + \frac{4}{3} ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{7}{6} ln ∣ 2 x + 1 ∣ + \frac{4}{3} ln ∣ x - 1 ∣ + C

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\frac{d}{d x} (lo g_{4} ((x^{2} + 2 x)^{3}))

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\int (1 + cos (y)) d y

What is the integral of (5x-1)/((2x+1)(x-1)) ?
The integral of (5x-1)/((2x+1)(x-1)) is 7/6 ln|2x+1|+4/3 ln|x-1|+C