What is the integral of (x^2-6)^4(2x) ?

The integral of (x^2-6)^4(2x) is 1/5 (x^2-6)^5+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2-6)^4(2x)

\int (x^{2} - 6)^{4} (2 x) d x

\frac{1}{5} (x^{2} - 6)^{5} + C

Solution steps

\int (x^{2} - 6)^{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} - 6)^{4} x 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} - 6

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 6 ) ^{5}}{5}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 6 ) ^{5}}{5} : \frac{1}{5} (x^{2} - 6)^{5}

= \frac{1}{5} (x^{2} - 6)^{5}

Add a constant to the solution

= \frac{1}{5} (x^{2} - 6)^{5} + C

x \to \infty lim (x (x - 1))

y = (5 x - 2)^{3}

\frac{x - 7}{x + 7}

x \to - 1 lim (3 - x^{2})

\int 2 sin (x) + 2 x^{2} + 5 x + 1 d x

What is the integral of (x^2-6)^4(2x) ?
The integral of (x^2-6)^4(2x) is 1/5 (x^2-6)^5+C