What is the integral from 1 to 2 of 1/(x(x+1)) ?

The integral from 1 to 2 of 1/(x(x+1)) is 2ln(2)-ln(3)

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integral from 1 to 2 of 1/(x(x+1))

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

2 ln (2) - ln (3)

Decimal Notation

0.28768 \dots

Solution steps

\int_{1}^{2} \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_{1}^{2} \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_{1}^{2} \frac{1}{x} d x - \int_{1}^{2} \frac{1}{x + 1} d x

\int_{1}^{2} \frac{1}{x} d x = ln (2)

\int_{1}^{2} \frac{1}{x + 1} d x = ln (3) - ln (2)

= ln (2) - (ln (3) - ln (2))

Simplify

= 2 ln (2) - ln (3)

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What is the integral from 1 to 2 of 1/(x(x+1)) ?
The integral from 1 to 2 of 1/(x(x+1)) is 2ln(2)-ln(3)