What is the integral of (x^2-9)^4(2x) ?

The integral of (x^2-9)^4(2x) is 1/5 (x^2-9)^5+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^2-9)^4(2x)

\int (x^{2} - 9)^{4} (2 x) d x

\frac{1}{5} (x^{2} - 9)^{5} + C

Solution steps

\int (x^{2} - 9)^{4} \cdot 2 x d x

Take the constant out:

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

= 2 \cdot \int (x^{2} - 9)^{4} x d x

Apply u-substitution

= 2 \cdot \int \frac{u ^{4}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int u^{4} d u

Apply the Power Rule

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{5}}{5}

Substitute back

u = x^{2} - 9

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 9 ) ^{5}}{5}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 9 ) ^{5}}{5} : \frac{1}{5} (x^{2} - 9)^{5}

= \frac{1}{5} (x^{2} - 9)^{5}

Add a constant to the solution

= \frac{1}{5} (x^{2} - 9)^{5} + C

What is the integral of (x^2-9)^4(2x) ?
The integral of (x^2-9)^4(2x) is 1/5 (x^2-9)^5+C