What is the integral from e to 6 of 1/(x(x+1)) ?

The integral from e to 6 of 1/(x(x+1)) is ln(6)-1-ln(7)+ln(e+1)

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integral from e to 6 of 1/(x(x+1))

\int_{e}^{6} \frac{1}{x ( x + 1 )} d x

ln (6) - 1 - ln (7) + ln (e + 1)

Decimal

0.15911 \dots

Solution steps

\int_{e}^{6} \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_{e}^{6} \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_{e}^{6} \frac{1}{x} d x - \int_{e}^{6} \frac{1}{x + 1} d x

\int_{e}^{6} \frac{1}{x} d x = ln (6) - 1

\int_{e}^{6} \frac{1}{x + 1} d x = ln (7) - ln (e + 1)

= ln (6) - 1 - (ln (7) - ln (e + 1))

Simplify

= ln (6) - 1 - ln (7) + ln (e + 1)

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What is the integral from e to 6 of 1/(x(x+1)) ?
The integral from e to 6 of 1/(x(x+1)) is ln(6)-1-ln(7)+ln(e+1)