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

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

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

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

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

Decimal

1.09816 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{e + 1} \frac{u - 1}{u} d u

= \int_{2}^{e + 1} \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \int_{2}^{e + 1} 1 - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{2}^{e + 1} 1 d u - \int_{2}^{e + 1} \frac{1}{u} d u

\int_{2}^{e + 1} 1 d u = e - 1

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

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

Simplify

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

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