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

The integral from 1 to 3 of (6x^2+6/(x^2)) is 56

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

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

56

Solution steps

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

\int_{1}^{3} 6 x^{2} d x = 52

\int_{1}^{3} \frac{6}{x ^{2}} d x = 4

= 52 + 4

Simplify

= 56

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What is the integral from 1 to 3 of (6x^2+6/(x^2)) ?
The integral from 1 to 3 of (6x^2+6/(x^2)) is 56