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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (\frac{x ^{2} y + 3 z y ^{3}}{x yz})

x y + y^{^{'}} = 80 x

\int_{3}^{4} (4 + \frac{24}{x ^{2}}) d x

\frac{d y}{d x} + 7 x y = 0

x \to 2 lim (x^{3} - x^{2} - 4)

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