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

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

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

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

- 2 ln (2) + \frac{1}{3}

Decimal

- 1.05296 \dots

Solution steps

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

Expand

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

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

\int_{1}^{2} x^{2} d x = \frac{7}{3}

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

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

= \frac{7}{3} - 2 - 2 ln (2)

Combine the fractions

- 2 + \frac{7}{3} : \frac{1}{3}

= - 2 ln (2) + \frac{1}{3}

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

\int_{- 1}^{2} 8 - x^{2} - x d x

\int_{- 2}^{2} 1 - \frac{x ^{2}}{4} d x

\int_{- 2}^{1} 2 x^{2} d x

\int_{- 1}^{1} (x^{2} - x) (x + 1) d x

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