What is the integral from 1 to 2 of (4x^2+8/(x^2)) ?

The integral from 1 to 2 of (4x^2+8/(x^2)) is 40/3

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integral from 1 to 2 of (4x^2+8/(x^2))

\int_{1}^{2} (4 x^{2} + \frac{8}{x ^{2}}) d x

\frac{40}{3}

Decimal Notation

13.33333 \dots

Solution steps

\int_{1}^{2} 4 x^{2} + \frac{8}{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}^{2} 4 x^{2} d x + \int_{1}^{2} \frac{8}{x ^{2}} d x

\int_{1}^{2} 4 x^{2} d x = \frac{28}{3}

\int_{1}^{2} \frac{8}{x ^{2}} d x = 4

= \frac{28}{3} + 4

Simplify

\frac{28}{3} + 4 : \frac{40}{3}

= \frac{40}{3}

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

\int_{0}^{1} (t - 1) d t

\int_{0}^{1} x^{\frac{1}{4}} - x^{\frac{1}{2}} d x

\int_{0}^{3} π (3 x - x^{2})^{2} d x

\int_{4}^{6} \frac{1}{x ^{2} - 9} d x

What is the integral from 1 to 2 of (4x^2+8/(x^2)) ?
The integral from 1 to 2 of (4x^2+8/(x^2)) is 40/3