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

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

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

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

1 - ln (e + 1) + ln (2)

Decimal

0.37988 \dots

Solution steps

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

Take the partial fraction of

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

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

\int_{1}^{e} \frac{1}{x} d x = 1

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

= 1 - (ln (e + 1) - ln (2))

Simplify

= 1 - ln (e + 1) + ln (2)

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