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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int e^{t y} d y

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + y = 0

\int \frac{x}{x + 7} d x

y^{^{''}} + 4 y = 4 sec (2 t)

x y^{^{'}} - y = 2020 x^{2}

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