What is the integral of (4x-1)^4 ?

The integral of (4x-1)^4 is 1/20 (4x-1)^5+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (4x-1)^4

\int (4 x - 1)^{4} d x

\frac{1}{20} (4 x - 1)^{5} + C

Solution steps

\int (4 x - 1)^{4} d x

Apply u-substitution

= \int \frac{u ^{4}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{4} d u

Apply the Power Rule

= \frac{1}{4} \cdot \frac{u ^{5}}{5}

Substitute back

u = 4 x - 1

= \frac{1}{4} \cdot \frac{( 4 x - 1 ) ^{5}}{5}

Simplify

\frac{1}{4} \cdot \frac{( 4 x - 1 ) ^{5}}{5} : \frac{1}{20} (4 x - 1)^{5}

= \frac{1}{20} (4 x - 1)^{5}

Add a constant to the solution

= \frac{1}{20} (4 x - 1)^{5} + C