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

The integral from 1 to 2 of (x^2+1/x) is 7/3+ln(2)

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

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

\frac{7}{3} + ln (2)

Decimal

3.02648 \dots

Solution steps

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

\int_{1}^{2} x^{2} d x = \frac{7}{3}

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

= \frac{7}{3} + ln (2)

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What is the integral from 1 to 2 of (x^2+1/x) ?
The integral from 1 to 2 of (x^2+1/x) is 7/3+ln(2)