What is the integral from 2 to 5 of ((3x^2+1))/x ?

The integral from 2 to 5 of ((3x^2+1))/x is 63/2+ln(5)-ln(2)

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integral from 2 to 5 of ((3x^2+1))/x

\int_{2}^{5} \frac{( 3 x ^{2} + 1 )}{x} d x

\frac{63}{2} + ln (5) - ln (2)

Decimal

32.41629 \dots

Solution steps

\int_{2}^{5} \frac{3 x ^{2} + 1}{x} d x

Expand

\frac{3 x ^{2} + 1}{x} : 3 x + \frac{1}{x}

= \int_{2}^{5} 3 x + \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} 3 x d x + \int_{2}^{5} \frac{1}{x} d x

\int_{2}^{5} 3 x d x = \frac{63}{2}

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

= \frac{63}{2} + ln (5) - ln (2)

What is the integral from 2 to 5 of ((3x^2+1))/x ?
The integral from 2 to 5 of ((3x^2+1))/x is 63/2+ln(5)-ln(2)