What is the integral of (x^2-4)^32x ?

The integral of (x^2-4)^32x is 1/4 (x^2-4)^4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2-4)^32x

\int (x^{2} - 4)^{3} 2 x d x

\frac{1}{4} (x^{2} - 4)^{4} + C

Solution steps

\int (x^{2} - 4)^{3} \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} - 4)^{3} x d x

Apply u-substitution

: \int \frac{u ^{3}}{2} d u

= 2 \cdot \int \frac{u ^{3}}{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^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

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

Substitute back

u = x^{2} - 4

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 4 ) ^{4}}{4}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 4 ) ^{4}}{4} : \frac{1}{4} (x^{2} - 4)^{4}

= \frac{1}{4} (x^{2} - 4)^{4}

Add a constant to the solution

= \frac{1}{4} (x^{2} - 4)^{4} + C

\frac{- 12}{( 1 - x ) ^{4}}

\int sinh (8 x) d x

\frac{d}{d x} (x x^{2} + 36)

\frac{d}{d x} (\frac{( x - x )}{x ^{\frac{1}{5}}})

100 - \frac{200}{t ^{2} + 4}

What is the integral of (x^2-4)^32x ?
The integral of (x^2-4)^32x is 1/4 (x^2-4)^4+C