What is the integral from 1 to 8 of ((x^2+3x+1)/x) ?

The integral from 1 to 8 of ((x^2+3x+1)/x) is 3ln(2)+105/2

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integral from 1 to 8 of ((x^2+3x+1)/x)

\int_{1}^{8} (\frac{x ^{2} + 3 x + 1}{x}) d x

3 ln (2) + \frac{105}{2}

Decimal Notation

54.57944 \dots

Solution steps

\int_{1}^{8} \frac{x ^{2} + 3 x + 1}{x} d x

Expand

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

= \int_{1}^{8} x + 3 + \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}^{8} x d x + \int_{1}^{8} 3 d x + \int_{1}^{8} \frac{1}{x} d x

\int_{1}^{8} x d x = \frac{63}{2}

\int_{1}^{8} 3 d x = 21

\int_{1}^{8} \frac{1}{x} d x = 3 ln (2)

= \frac{63}{2} + 21 + 3 ln (2)

Combine the fractions

21 + \frac{63}{2} : \frac{105}{2}

= 3 ln (2) + \frac{105}{2}

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What is the integral from 1 to 8 of ((x^2+3x+1)/x) ?
The integral from 1 to 8 of ((x^2+3x+1)/x) is 3ln(2)+105/2