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

The integral of x^2(4-x^3)^4 is 1/15 (-4+x^3)^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(4-x^3)^4

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

\frac{1}{15} (- 4 + x^{3})^{5} + C

Solution steps

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

Factor

4 - x^{3} : - (- 4 + x^{3})

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

(- (- 4 + x^{3}))^{4} = (- 4 + x^{3})^{4}

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

Apply u-substitution

: \int \frac{u ^{4}}{3} d u

= \int \frac{u ^{4}}{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^{4} d u

Apply the Power Rule

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

= \frac{1}{3} \cdot \frac{u ^{5}}{5}

Substitute back

u = - 4 + x^{3}

= \frac{1}{3} \cdot \frac{( - 4 + x ^{3} ) ^{5}}{5}

Simplify

\frac{1}{3} \cdot \frac{( - 4 + x ^{3} ) ^{5}}{5} : \frac{1}{15} (- 4 + x^{3})^{5}

= \frac{1}{15} (- 4 + x^{3})^{5}

Add a constant to the solution

= \frac{1}{15} (- 4 + x^{3})^{5} + C

\int (x^{2} + x) (e^{1 - 2 x}) d x

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

\int cot^{3} (y) csc^{2} (y) d y

\int \frac{3 x}{( x ^{2} - 1 ) ^{5}} d x

What is the integral of x^2(4-x^3)^4 ?
The integral of x^2(4-x^3)^4 is 1/15 (-4+x^3)^5+C