What is the integral from 1 to 3 of ((x^2+4)/x) ?

The integral from 1 to 3 of ((x^2+4)/x) is 4+4ln(3)

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integral from 1 to 3 of ((x^2+4)/x)

\int_{1}^{3} (\frac{x ^{2} + 4}{x}) d x

4 + 4 ln (3)

Decimal

8.39444 \dots

Solution steps

\int_{1}^{3} \frac{x ^{2} + 4}{x} d x

Expand

\frac{x ^{2} + 4}{x} : x + \frac{4}{x}

= \int_{1}^{3} x + \frac{4}{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}^{3} x d x + \int_{1}^{3} \frac{4}{x} d x

\int_{1}^{3} x d x = 4

\int_{1}^{3} \frac{4}{x} d x = 4 ln (3)

= 4 + 4 ln (3)

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What is the integral from 1 to 3 of ((x^2+4)/x) ?
The integral from 1 to 3 of ((x^2+4)/x) is 4+4ln(3)