What is the integral from 2 to 5 of (x^2+1/x) ?

The integral from 2 to 5 of (x^2+1/x) is 39+ln(5)-ln(2)

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integral from 2 to 5 of (x^2+1/x)

\int_{2}^{5} (x^{2} + \frac{1}{x}) d x

39 + ln (5) - ln (2)

Decimal

39.91629 \dots

Solution steps

\int_{2}^{5} 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_{2}^{5} x^{2} d x + \int_{2}^{5} \frac{1}{x} d x

\int_{2}^{5} x^{2} d x = 39

\int_{2}^{5} \frac{1}{x} d x = ln (5) - ln (2)

= 39 + ln (5) - ln (2)

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What is the integral from 2 to 5 of (x^2+1/x) ?
The integral from 2 to 5 of (x^2+1/x) is 39+ln(5)-ln(2)