What is the integral of x(x^2-1)^6 ?

The integral of x(x^2-1)^6 is 1/14 (x^2-1)^7+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x(x^2-1)^6

\int x (x^{2} - 1)^{6} d x

\frac{1}{14} (x^{2} - 1)^{7} + C

Solution steps

\int x (x^{2} - 1)^{6} d x

Apply u-substitution

= \int \frac{u ^{6}}{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^{6} d u

Apply the Power Rule

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

Substitute back

u = x^{2} - 1

= \frac{1}{2} \cdot \frac{( x ^{2} - 1 ) ^{7}}{7}

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} - 1 ) ^{7}}{7} : \frac{1}{14} (x^{2} - 1)^{7}

= \frac{1}{14} (x^{2} - 1)^{7}

Add a constant to the solution

= \frac{1}{14} (x^{2} - 1)^{7} + C

What is the integral of x(x^2-1)^6 ?
The integral of x(x^2-1)^6 is 1/14 (x^2-1)^7+C