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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6x^4-5x^3+5x^2+C

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

\frac{6 x ^{5}}{5} - \frac{5 x ^{4}}{4} + \frac{5 x ^{3}}{3} + C x + C

Solution steps

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

\int 6 x^{4} d x = \frac{6 x ^{5}}{5}

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

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

\int C d x = C x

= \frac{6 x ^{5}}{5} - \frac{5 x ^{4}}{4} + \frac{5 x ^{3}}{3} + C x

Add a constant to the solution

= \frac{6 x ^{5}}{5} - \frac{5 x ^{4}}{4} + \frac{5 x ^{3}}{3} + C x + C

\frac{d y}{d x} = e^{2 y - x}

y = (7 x - 4) (2 + 6 x)

x \to \infty lim (arctan (1 - x))

x \to 0 + lim (arctan (ln (x)))

\frac{d v}{d x} - \frac{1}{x} v = 8

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