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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

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

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (x^{3} + x - 1)

2 e^{(x - 5)}

\frac{x - 4}{x + 4}

\frac{d y}{d x} = 2 y

x \to 0 lim (sin (\frac{40 π}{x}))

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