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

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

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

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

- 2

Solution steps

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

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

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

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

Simplify

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

= - 2

\int \frac{1}{x ^{2} + 3 x + 9} d x

f (x) = \frac{1}{x + 2}

\frac{\partial}{\partial y} (y^{3} + x y^{2} + y)

\int \frac{x ^{2}}{x ^{6} + 9} d x

\frac{\partial}{\partial x} (lo g_{e} (x^{2} + y^{2}))

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