What is the integral from 1 to 3 of 3/(x(x+1)) ?

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

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integral from 1 to 3 of 3/(x(x+1))

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

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

Decimal

1.21639 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int_{1}^{3} \frac{1}{x ( x + 1 )} d x

Take the partial fraction of

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

= 3 \cdot \int_{1}^{3} \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

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

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

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

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

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