What is the integral from 6 to 9 of (x^2+1/x) ?

The integral from 6 to 9 of (x^2+1/x) is 171+ln(3/2)

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integral from 6 to 9 of (x^2+1/x)

\int_{6}^{9} (x^{2} + \frac{1}{x}) d x

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

Decimal

171.40546 \dots

Solution steps

\int_{6}^{9} 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_{6}^{9} x^{2} d x + \int_{6}^{9} \frac{1}{x} d x

\int_{6}^{9} x^{2} d x = 171

\int_{6}^{9} \frac{1}{x} d x = ln (\frac{3}{2})

= 171 + ln (\frac{3}{2})

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What is the integral from 6 to 9 of (x^2+1/x) ?
The integral from 6 to 9 of (x^2+1/x) is 171+ln(3/2)