What is the integral of x^2(1-x^3)^4 ?

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

Popular >

integral of x^2(1-x^3)^4

\int x^{2} (1 - x^{3})^{4} d x

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

Solution steps

\int x^{2} (1 - 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 = 1 - x^{3}

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

Simplify

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

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

Add a constant to the solution

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

\frac{s ^{2} - 2 s + 3}{( s - 2 ) ^{3}}

\int \frac{5 x + 3}{x ^{2} - 9} d x

\int x^{3} sin (6 x) d x

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

\frac{d}{d x} (1 - \frac{1}{1 + x})

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