What is the integral of (2x-1)^7 ?

The integral of (2x-1)^7 is 1/16 (2x-1)^8+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x-1)^7

\int (2 x - 1)^{7} d x

\frac{1}{16} (2 x - 1)^{8} + C

Solution steps

\int (2 x - 1)^{7} d x

Apply u-substitution

: \int \frac{u ^{7}}{2} d u

= \int \frac{u ^{7}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{7} d u

Apply the Power Rule

: \frac{u ^{8}}{8}

= \frac{1}{2} \cdot \frac{u ^{8}}{8}

Substitute back

u = 2 x - 1

= \frac{1}{2} \cdot \frac{( 2 x - 1 ) ^{8}}{8}

Simplify

\frac{1}{2} \cdot \frac{( 2 x - 1 ) ^{8}}{8} : \frac{1}{16} (2 x - 1)^{8}

= \frac{1}{16} (2 x - 1)^{8}

Add a constant to the solution

= \frac{1}{16} (2 x - 1)^{8} + C

\int \frac{1}{2 v} d v

\int \frac{x}{2 x + 1} d x

\int (2 x^{2} + 4)^{8} \cdot x d x

\int \frac{sin ( 3 x )}{e ^{x}} d x