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

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

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

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

e

Decimal

2.71828 \dots

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{u + 1}{u} : 1 + \frac{1}{u}

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

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

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

= e - 1 + 1

Simplify

= e

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