What is the integral of x^2(x^3-4)^3 ?

The integral of x^2(x^3-4)^3 is 1/12 (x^3-4)^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(x^3-4)^3

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

\frac{1}{12} (x^{3} - 4)^{4} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{3} \cdot \int u^{3} d u

Apply the Power Rule

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

= \frac{1}{3} \cdot \frac{u ^{4}}{4}

Substitute back

u = x^{3} - 4

= \frac{1}{3} \cdot \frac{( x ^{3} - 4 ) ^{4}}{4}

Simplify

\frac{1}{3} \cdot \frac{( x ^{3} - 4 ) ^{4}}{4} : \frac{1}{12} (x^{3} - 4)^{4}

= \frac{1}{12} (x^{3} - 4)^{4}

Add a constant to the solution

= \frac{1}{12} (x^{3} - 4)^{4} + C

\frac{d}{d x} (\frac{6 x}{x ^{2} + 3})

\frac{d}{d x} (\frac{x - 7}{( x + 5 ) ^{10}})

x^{2} - 12 x + k

x \to 4 lim (x 25 - x^{2})

y^{^{''}} + 4 y^{^{'}} + 12 y = 0

What is the integral of x^2(x^3-4)^3 ?
The integral of x^2(x^3-4)^3 is 1/12 (x^3-4)^4+C