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

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

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

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

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

Solution steps

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

Expand

\frac{5 x - 1}{x ^{2} ( 2 x - 1 )} : \frac{5 x}{x ^{2} ( 2 x - 1 )} - \frac{1}{x ^{2} ( 2 x - 1 )}

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

\int \frac{5 x}{x ^{2} ( 2 x - 1 )} d x = 5 (- ln ∣ x ∣ + ln ∣ 2 x - 1 ∣)

\int \frac{1}{x ^{2} ( 2 x - 1 )} d x = - 2 ln ∣ x ∣ + \frac{1}{x} + 2 ln ∣ 2 x - 1 ∣

= 5 (- ln ∣ x ∣ + ln ∣ 2 x - 1 ∣) - (- 2 ln ∣ x ∣ + \frac{1}{x} + 2 ln ∣ 2 x - 1 ∣)

Simplify

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

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

Add a constant to the solution

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

\frac{s + 2}{2 s \cdot ( s + 1 )}

\frac{d ^{2}}{d x ^{2}} (x y e^{(- x + 3 y)})

\frac{\partial}{\partial z} (3 x y z^{2})

e^{2 x} ln (2 x)

x \to - 2 lim (\frac{8}{x})

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