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

The integral from 1 to 2 of ((3x^2+9x+1)/x) is ln(2)+27/2

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integral from 1 to 2 of ((3x^2+9x+1)/x)

\int_{1}^{2} (\frac{3 x ^{2} + 9 x + 1}{x}) d x

ln (2) + \frac{27}{2}

Decimal

14.19314 \dots

Solution steps

\int_{1}^{2} \frac{3 x ^{2} + 9 x + 1}{x} d x

Expand

\frac{3 x ^{2} + 9 x + 1}{x} : 3 x + 9 + \frac{1}{x}

= \int_{1}^{2} 3 x + 9 + \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_{1}^{2} 3 x d x + \int_{1}^{2} 9 d x + \int_{1}^{2} \frac{1}{x} d x

\int_{1}^{2} 3 x d x = \frac{9}{2}

\int_{1}^{2} 9 d x = 9

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

= \frac{9}{2} + 9 + ln (2)

Combine the fractions

9 + \frac{9}{2} : \frac{27}{2}

= ln (2) + \frac{27}{2}

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What is the integral from 1 to 2 of ((3x^2+9x+1)/x) ?
The integral from 1 to 2 of ((3x^2+9x+1)/x) is ln(2)+27/2