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

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

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

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

- 2

Solution steps

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

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

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

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

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

Simplify

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

= - 2

1 + x^{2}

\frac{d y}{d x} = e^{2 x} (1 + y^{2})

y = 3 x^{2} + 5

\int \frac{2}{5 cos ^{2} ( x )} d x

\int sec (2 x) tan^{3} (2 x) d x

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