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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int ln (x^{2} - 81) d x

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

\int \frac{x + ln ( x ^{2} )}{x} d x

\int \frac{1}{( 3 x )} d x

\int \frac{10}{x ln ( x )} d 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