What is the integral from 3 to 4.3 of (12x^2-6x+4/x) ?

The integral from 3 to 4.3 of (12x^2-6x+4/x) is 182.99801…

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integral from 3 to 4.3 of (12x^2-6x+4/x)

\int_{3}^{4.3} (12 x^{2} - 6 x + \frac{4}{x}) d x

182.99801 \dots

Solution steps

\int_{3}^{4.3} 12 x^{2} - 6 x + \frac{4}{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_{3}^{4.3} 12 x^{2} d x - \int_{3}^{4.3} 6 x d x + \int_{3}^{4.3} \frac{4}{x} d x

\int_{3}^{4.3} 12 x^{2} d x = 210.028

\int_{3}^{4.3} 6 x d x = 28.47

\int_{3}^{4.3} \frac{4}{x} d x = 1.44001 \dots

= 210.028 - 28.47 + 1.44001 \dots

Simplify

= 182.99801 \dots

What is the integral from 3 to 4.3 of (12x^2-6x+4/x) ?
The integral from 3 to 4.3 of (12x^2-6x+4/x) is 182.99801…