What is the integral from 1 to 2 of (9x^3+4x^2-2)/x ?

The integral from 1 to 2 of (9x^3+4x^2-2)/x is -2ln(2)+27

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integral from 1 to 2 of (9x^3+4x^2-2)/x

\int_{1}^{2} \frac{9 x ^{3} + 4 x ^{2} - 2}{x} d x

- 2 ln (2) + 27

Decimal

25.61370 \dots

Solution steps

\int_{1}^{2} \frac{9 x ^{3} + 4 x ^{2} - 2}{x} d x

Expand

\frac{9 x ^{3} + 4 x ^{2} - 2}{x} : 9 x^{2} + 4 x - \frac{2}{x}

= \int_{1}^{2} 9 x^{2} + 4 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}^{2} 9 x^{2} d x + \int_{1}^{2} 4 x d x - \int_{1}^{2} \frac{2}{x} d x

\int_{1}^{2} 9 x^{2} d x = 21

\int_{1}^{2} 4 x d x = 6

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

= 21 + 6 - 2 ln (2)

Simplify

= - 2 ln (2) + 27

What is the integral from 1 to 2 of (9x^3+4x^2-2)/x ?
The integral from 1 to 2 of (9x^3+4x^2-2)/x is -2ln(2)+27