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

The integral from 1 to e of (x^2+1)/(2x) is (e^2+1)/4

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

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

\frac{e ^{2} + 1}{4}

Decimal

2.09726 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

\frac{x ^{2} + 1}{x} : x + \frac{1}{x}

= \frac{1}{2} \cdot \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

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

Simplify

\frac{1}{2} (\frac{e ^{2} - 1}{2} + 1) : \frac{e ^{2} + 1}{4}

= \frac{e ^{2} + 1}{4}

\int_{0}^{7} 4 d x

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\int_{- 1}^{1} (\frac{6}{x ^{2} + 1}) d x

What is the integral from 1 to e of (x^2+1)/(2x) ?
The integral from 1 to e of (x^2+1)/(2x) is (e^2+1)/4