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

The integral of 3/4+4/5 x^2-5/6 x^3 is 3/4 x+(4x^3)/(15)-(5x^4)/(24)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{3}{4} x + \frac{4 x ^{3}}{15} - \frac{5 x ^{4}}{24} + C

Solution steps

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

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

\int \frac{4}{5} x^{2} d x = \frac{4 x ^{3}}{15}

\int \frac{5}{6} x^{3} d x = \frac{5 x ^{4}}{24}

= \frac{3}{4} x + \frac{4 x ^{3}}{15} - \frac{5 x ^{4}}{24}

Add a constant to the solution

= \frac{3}{4} x + \frac{4 x ^{3}}{15} - \frac{5 x ^{4}}{24} + C

5 x + 7 x

\frac{d}{d x} (x^{2} (x^{2} - 4))

\frac{d}{d x} (\frac{3}{5} x^{\frac{5}{3}})

1 + x

What is the integral of 3/4+4/5 x^2-5/6 x^3 ?
The integral of 3/4+4/5 x^2-5/6 x^3 is 3/4 x+(4x^3)/(15)-(5x^4)/(24)+C