What is the integral of (3x^2+4x-5) ?

The integral of (3x^2+4x-5) is x^3+2x^2-5x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (3x^2+4x-5)

\int (3 x^{2} + 4 x - 5) d x

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

Solution steps

\int 3 x^{2} + 4 x - 5 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 3 x^{2} d x + \int 4 x d x - \int 5 d x

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

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

\int 5 d x = 5 x

= x^{3} + 2 x^{2} - 5 x

Add a constant to the solution

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

What is the integral of (3x^2+4x-5) ?
The integral of (3x^2+4x-5) is x^3+2x^2-5x+C