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

The integral of x(1-x)^3 is (x^2)/2-x^3+(3x^4)/4-(x^5)/5+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x(1-x)^3

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

\frac{x ^{2}}{2} - x^{3} + \frac{3 x ^{4}}{4} - \frac{x ^{5}}{5} + C

Solution steps

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

Expand

x (1 - x)^{3} : x - 3 x^{2} + 3 x^{3} - x^{4}

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

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

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

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

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

= \frac{x ^{2}}{2} - x^{3} + \frac{3 x ^{4}}{4} - \frac{x ^{5}}{5}

Add a constant to the solution

= \frac{x ^{2}}{2} - x^{3} + \frac{3 x ^{4}}{4} - \frac{x ^{5}}{5} + C

\int x tan^{2} (\frac{x}{2}) d x

\int - tan (x) cos (x) d x

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

\int \frac{arccos ^{2} ( x )}{1 - x ^{2}} d x

\int 3 x^{2} + 4 x - 7 d x

What is the integral of x(1-x)^3 ?
The integral of x(1-x)^3 is (x^2)/2-x^3+(3x^4)/4-(x^5)/5+C