What is the integral from 1 to 6 of ((3x^2+2x+1)/x) ?

The integral from 1 to 6 of ((3x^2+2x+1)/x) is ln(6)+125/2

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integral from 1 to 6 of ((3x^2+2x+1)/x)

\int_{1}^{6} (\frac{3 x ^{2} + 2 x + 1}{x}) d x

ln (6) + \frac{125}{2}

Decimal

64.29175 \dots

Solution steps

\int_{1}^{6} \frac{3 x ^{2} + 2 x + 1}{x} d x

Expand

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

= \int_{1}^{6} 3 x + 2 + \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}^{6} 3 x d x + \int_{1}^{6} 2 d x + \int_{1}^{6} \frac{1}{x} d x

\int_{1}^{6} 3 x d x = \frac{105}{2}

\int_{1}^{6} 2 d x = 10

\int_{1}^{6} \frac{1}{x} d x = ln (6)

= \frac{105}{2} + 10 + ln (6)

Combine the fractions

10 + \frac{105}{2} : \frac{125}{2}

= ln (6) + \frac{125}{2}

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What is the integral from 1 to 6 of ((3x^2+2x+1)/x) ?
The integral from 1 to 6 of ((3x^2+2x+1)/x) is ln(6)+125/2