What is the integral from 1 to 2 of (x^2+x^{-4}) ?

The integral from 1 to 2 of (x^2+x^{-4}) is 21/8

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integral from 1 to 2 of (x^2+x^{-4})

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

\frac{21}{8}

Decimal

2.625

Solution steps

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

\int_{1}^{2} x^{2} d x = \frac{7}{3}

\int_{1}^{2} x^{- 4} d x = \frac{7}{24}

= \frac{7}{3} + \frac{7}{24}

Simplify

\frac{7}{3} + \frac{7}{24} : \frac{21}{8}

= \frac{21}{8}

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What is the integral from 1 to 2 of (x^2+x^{-4}) ?
The integral from 1 to 2 of (x^2+x^{-4}) is 21/8