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

The integral of (1-x)^n is -((1-x)^{n+1})/(n+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1-x)^n

integral

(1 - x)^{n}

- \frac{( 1 - x ) ^{n + 1}}{n + 1} + C

Solution steps

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

Apply u-substitution

: \int - u^{n} d u

= \int - u^{n} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int u^{n} d u

Apply the Power Rule

: \frac{u ^{n + 1}}{n + 1}

= - \frac{u ^{n + 1}}{n + 1}

Substitute back

u = 1 - x

= - \frac{( 1 - x ) ^{n + 1}}{n + 1}

Add a constant to the solution

= - \frac{( 1 - x ) ^{n + 1}}{n + 1} + C

\frac{1}{2 sin ( 2 x )}

(a^{2} + x^{2})^{\frac{1}{2}}

x^{4} (1 - x)^{5}

\frac{x ^{3}}{( x ^{2} + x + 2 )}

(x^{2} + 1)^{3}

What is the integral of (1-x)^n ?
The integral of (1-x)^n is -((1-x)^{n+1})/(n+1)+C