What is the integral of (x^3-3x^2+2x+1) ?

The integral of (x^3-3x^2+2x+1) is (x^4)/4-x^3+x^2+x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^3-3x^2+2x+1)

\int (x^{3} - 3 x^{2} + 2 x + 1) d x

\frac{x ^{4}}{4} - x^{3} + x^{2} + x + C

Solution steps

\int x^{3} - 3 x^{2} + 2 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 x^{3} d x - \int 3 x^{2} d x + \int 2 x d x + \int 1 d x

\int x^{3} d x = \frac{x ^{4}}{4}

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

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

\int 1 d x = x

= \frac{x ^{4}}{4} - x^{3} + x^{2} + x

Add a constant to the solution

= \frac{x ^{4}}{4} - x^{3} + x^{2} + x + C

What is the integral of (x^3-3x^2+2x+1) ?
The integral of (x^3-3x^2+2x+1) is (x^4)/4-x^3+x^2+x+C