What is the integral of 2x(x^2+9)^5 ?

The integral of 2x(x^2+9)^5 is 1/6 (x^2+9)^6+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 2x(x^2+9)^5

\int 2 x (x^{2} + 9)^{5} d x

\frac{1}{6} (x^{2} + 9)^{6} + C

Solution steps

\int 2 x (x^{2} + 9)^{5} d x

Take the constant out:

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

= 2 \cdot \int x (x^{2} + 9)^{5} d x

Apply u-substitution

= 2 \cdot \int \frac{u ^{5}}{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^{5} d u

Apply the Power Rule

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = x^{2} + 9

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 9 ) ^{6}}{6}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 9 ) ^{6}}{6} : \frac{1}{6} (x^{2} + 9)^{6}

= \frac{1}{6} (x^{2} + 9)^{6}

Add a constant to the solution

= \frac{1}{6} (x^{2} + 9)^{6} + C

What is the integral of 2x(x^2+9)^5 ?
The integral of 2x(x^2+9)^5 is 1/6 (x^2+9)^6+C