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

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

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

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

2 ln (3) + 2

Decimal

4.19722 \dots

Solution steps

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

Expand

\frac{3 x ^{3} - 6 x ^{2} + 2}{x} : 3 x^{2} - 6 x + \frac{2}{x}

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

\int_{1}^{3} 3 x^{2} d x = 26

\int_{1}^{3} 6 x d x = 24

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

= 26 - 24 + 2 ln (3)

Simplify

= 2 ln (3) + 2

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