What is the integral of (6x^3-3\sqrt[3]{x})/x ?

The integral of (6x^3-3\sqrt[3]{x})/x is 2x^3-9\sqrt[3]{x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6x^3-3\sqrt[3]{x})/x

Solution steps

Expand

= \int 6 x^{2} - \frac{3}{x ^{\frac{2}{3}}} 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 6 x^{2} d x - \int \frac{3}{x ^{\frac{2}{3}}} d x

\int 6 x^{2} d x = 2 x^{3}

\int \frac{3}{x ^{\frac{2}{3}}} d x = 9 x^{\frac{1}{3}}

= 2 x^{3} - 9 x^{\frac{1}{3}}

Apply radical rule:

Add a constant to the solution

\int 8 x^{7} e^{8 x^{8}} d x

\int \frac{x - 2}{( x - 1 ) ( x ^{2} - 2 x + 5 )} d x

\int \frac{x}{( x + 2 ) ^{\frac{1}{4}}} d x

\int \frac{( x - 1 )}{x ^{2} - 2 x} d x

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

What is the integral of (6x^3-3\sqrt[3]{x})/x ?
The integral of (6x^3-3\sqrt[3]{x})/x is 2x^3-9\sqrt[3]{x}+C