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

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

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

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

- ln ∣ x ∣ + 2 ln ∣ x + 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= - ln ∣ x ∣ + 2 ln ∣ x + 1 ∣

Add a constant to the solution

= - ln ∣ x ∣ + 2 ln ∣ x + 1 ∣ + C

\frac{d}{d x} ((4 x^{3} + 1)^{2})

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\frac{d}{d x} (6 cos (π x))

x \to + 0 lim (\frac{sin ( 4 x )}{tan ( 2 x )})

f (x) = a x + b

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