What is the integral of (6x^2+3x)/(3x) ?

The integral of (6x^2+3x)/(3x) is x(x+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (6x^2+3x)/(3x)

\int \frac{6 x ^{2} + 3 x}{3 x} d x

x (x + 1) + C

Solution steps

\int \frac{6 x ^{2} + 3 x}{3 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{6 x ^{2} + 3 x}{x} d x

Expand

\frac{6 x ^{2} + 3 x}{x} : \frac{6 x ^{2}}{x} + \frac{3 x}{x}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int \frac{6 x ^{2}}{x} d x + \int \frac{3 x}{x} d x)

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

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

= \frac{1}{3} (3 x^{2} + 3 x)

Simplify

\frac{1}{3} (3 x^{2} + 3 x) : x (x + 1)

= x (x + 1)

Add a constant to the solution

= x (x + 1) + C

What is the integral of (6x^2+3x)/(3x) ?
The integral of (6x^2+3x)/(3x) is x(x+1)+C