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

The integral of (x^3-8)/(x-2) is (x^3)/3+x^2+4x+C

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integral of (x^3-8)/(x-2)

\int \frac{x ^{3} - 8}{x - 2} d x

\frac{x ^{3}}{3} + x^{2} + 4 x + C

Solution steps

\int \frac{x ^{3} - 8}{x - 2} d x

Simplify

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

= \int 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 x^{2} d x + \int 2 x d x + \int 4 d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 2 x d x = x^{2}

\int 4 d x = 4 x

= \frac{x ^{3}}{3} + x^{2} + 4 x

Add a constant to the solution

= \frac{x ^{3}}{3} + x^{2} + 4 x + C

\frac{5}{12} (3 + 5) a^{3}

\int x^{2} + y^{5} d y

\frac{d}{d x} (\frac{x ^{2} - 2 e ^{x}}{x + 3 e ^{x}})

y = 7 x + 11, (2, 5)

\frac{lo g _{6} ( x )}{e ^{4 x}}

What is the integral of (x^3-8)/(x-2) ?
The integral of (x^3-8)/(x-2) is (x^3)/3+x^2+4x+C