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

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

Popular >

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u ^{4}}{2} d u

= 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

: \frac{u ^{5}}{5}

= 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

\int \frac{6 x - 1}{( 2 x - 1 ) x ^{3}} d x

\int \frac{x ^{2} + 2 x + 4}{x ^{3} + 3 x ^{2} + 3 x + 1} d x

\int \frac{t ^{2} + 2 t - 1}{2 t ^{3} + 3 t ^{2} - 2 t} d t

\int \frac{1}{8 - 4 x} d x

\int \frac{x ^{3}}{( x ^{4} - 8 ) ^{4}} d x

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