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

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

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

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

- 1

Solution steps

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

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

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

= \frac{1}{2} - \frac{3}{2}

Simplify

\frac{1}{2} - \frac{3}{2} : - 1

= - 1

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