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

The integral from 1 to 2 of (1/x+x) is ln(2)+3/2

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

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

ln (2) + \frac{3}{2}

Decimal Notation

2.19314 \dots

Solution steps

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

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

\int_{1}^{2} x d x = \frac{3}{2}

= ln (2) + \frac{3}{2}

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What is the integral from 1 to 2 of (1/x+x) ?
The integral from 1 to 2 of (1/x+x) is ln(2)+3/2