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

The integral from 1 to 4 of (3x^2+4/(x^2)) is 66

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

\int_{1}^{4} (3 x^{2} + \frac{4}{x ^{2}}) d x

66

Solution steps

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

\int_{1}^{4} 3 x^{2} d x = 63

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

= 63 + 3

Simplify

= 66

\int_{1}^{4} (\frac{1}{x}) d x

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What is the integral from 1 to 4 of (3x^2+4/(x^2)) ?
The integral from 1 to 4 of (3x^2+4/(x^2)) is 66