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

The integral from-1 to 1 of ((x^3-8)/(x-2)) is 26/3

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

\int_{- 1}^{1} (\frac{x ^{3} - 8}{x - 2}) d x

\frac{26}{3}

Decimal Notation

8.66666 \dots

Solution steps

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

Simplify

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

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

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

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

\int_{- 1}^{1} 4 d x = 8

= \frac{2}{3} + 0 + 8

Simplify

\frac{2}{3} + 0 + 8 : \frac{26}{3}

= \frac{26}{3}

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

\int_{0.5}^{0} x d x

\int_{2}^{5} 6 x^{- 2} d x

\int_{4}^{7} e^{2 x + 1} d x

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

What is the integral from-1 to 1 of ((x^3-8)/(x-2)) ?
The integral from-1 to 1 of ((x^3-8)/(x-2)) is 26/3