What is the integral from 0 to-1 of (x^3+8)/(x+2) ?

The integral from 0 to-1 of (x^3+8)/(x+2) is -16/3

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integral from 0 to-1 of (x^3+8)/(x+2)

\int_{0}^{- 1} \frac{x ^{3} + 8}{x + 2} d x

- \frac{16}{3}

Decimal Notation

- 5.33333 \dots

Solution steps

\int_{0}^{- 1} \frac{x ^{3} + 8}{x + 2} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{0} \frac{x ^{3} + 8}{x + 2} d x

Simplify

\frac{x ^{3} + 8}{x + 2} : x^{2} - 2 x + 4

= - \int_{- 1}^{0} x^{2} - 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}^{0} x^{2} d x - \int_{- 1}^{0} 2 x d x + \int_{- 1}^{0} 4 d x)

\int_{- 1}^{0} x^{2} d x = \frac{1}{3}

\int_{- 1}^{0} 2 x d x = - 1

\int_{- 1}^{0} 4 d x = 4

= - (\frac{1}{3} - (- 1) + 4)

Simplify

- (\frac{1}{3} - (- 1) + 4) : - \frac{16}{3}

= - \frac{16}{3}

\frac{d y}{d x} + \frac{y}{x} = x^{3} y^{2}

\frac{d}{d x} (ln e^{1 + x^{2}} + 3 x^{4} + cot^{2} (π x))

(\frac{1}{1 + e ^{- x}})^{^{'}}

6 x^{\frac{3}{5}}

\frac{tan ( x )}{x}

What is the integral from 0 to-1 of (x^3+8)/(x+2) ?
The integral from 0 to-1 of (x^3+8)/(x+2) is -16/3