What is the integral of (6x^2-5x-1) ?

The integral of (6x^2-5x-1) is 2x^3-(5x^2)/2-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^2-5x-1)

\int (6 x^{2} - 5 x - 1) d x

2 x^{3} - \frac{5 x ^{2}}{2} - x + C

Solution steps

\int 6 x^{2} - 5 x - 1 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 5 x d x - \int 1 d x

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

\int 5 x d x = \frac{5 x ^{2}}{2}

\int 1 d x = x

= 2 x^{3} - \frac{5 x ^{2}}{2} - x

Add a constant to the solution

= 2 x^{3} - \frac{5 x ^{2}}{2} - x + C

x \to \infty lim (\frac{i}{n})

\frac{d y}{d x} - y - \frac{y ^{2}}{4} = 0

\int (2 x + 8 x^{2}) d x

\frac{d}{d x} (\frac{y}{x ^{2}})

y = x^{2} - 4, y = 0

What is the integral of (6x^2-5x-1) ?
The integral of (6x^2-5x-1) is 2x^3-(5x^2)/2-x+C