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

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

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

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

\frac{3}{2} + 4 ln (2)

Solution steps

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

Expand

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

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

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

\int_{1}^{2} \frac{4}{x} d x = 4 ln (2)

= \frac{3}{2} + 4 ln (2)

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