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

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

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

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

\frac{e ^{2} - 1}{2} + 1

Decimal

4.19452 \dots

Solution steps

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

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

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

= \frac{e ^{2} - 1}{2} + 1

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