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

The integral from-2 to-1 of (x^2+1/(x^2)) is 17/6

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

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

\frac{17}{6}

Decimal

2.83333 \dots

Solution steps

\int_{- 2}^{- 1} x^{2} + \frac{1}{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_{- 2}^{- 1} x^{2} d x + \int_{- 2}^{- 1} \frac{1}{x ^{2}} d x

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

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

= \frac{7}{3} + \frac{1}{2}

Simplify

\frac{7}{3} + \frac{1}{2} : \frac{17}{6}

= \frac{17}{6}

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What is the integral from-2 to-1 of (x^2+1/(x^2)) ?
The integral from-2 to-1 of (x^2+1/(x^2)) is 17/6