What is the integral of (3x^4-5x^2+6x) ?

The integral of (3x^4-5x^2+6x) is (3x^5)/5-(5x^3)/3+3x^2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (3x^4-5x^2+6x)

\int (3 x^{4} - 5 x^{2} + 6 x) d x

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

Solution steps

\int 3 x^{4} - 5 x^{2} + 6 x 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^{4} d x - \int 5 x^{2} d x + \int 6 x d x

\int 3 x^{4} d x = \frac{3 x ^{5}}{5}

\int 5 x^{2} d x = \frac{5 x ^{3}}{3}

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

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

Add a constant to the solution

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

What is the integral of (3x^4-5x^2+6x) ?
The integral of (3x^4-5x^2+6x) is (3x^5)/5-(5x^3)/3+3x^2+C