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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{cosh ( x )}{sin ^{2} ( x )})

\frac{d}{d x} (5 cos (5 x))

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

\int \frac{1}{x ( x + 1 )} d x

y = 2 - x^{2}

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