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

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

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

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

ln ∣ x ∣ - ln ∣ x + 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - ln ∣ x + 1 ∣

Add a constant to the solution

= ln ∣ x ∣ - ln ∣ x + 1 ∣ + C

x \to 0 lim (\frac{( x + 5 )}{3 x})

y = (t^{2} + 3) e^{t}

y = 3 + 4 x^{2} - 2 x^{3}

y^{^{''}} + 9 y = t sin (5 t)

\frac{d}{d x} (5 5 x - 4)

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