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

The integral of x^2(2x^3-1)^4 is 1/30 (2x^3-1)^5+C

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

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

\frac{1}{30} (2 x^{3} - 1)^{5} + C

Solution steps

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

Apply u-substitution

= \int \frac{u ^{4}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int u^{4} d u

Apply the Power Rule

= \frac{1}{6} \cdot \frac{u ^{5}}{5}

Substitute back

u = 2 x^{3} - 1

= \frac{1}{6} \cdot \frac{( 2 x ^{3} - 1 ) ^{5}}{5}

Simplify

\frac{1}{6} \cdot \frac{( 2 x ^{3} - 1 ) ^{5}}{5} : \frac{1}{30} (2 x^{3} - 1)^{5}

= \frac{1}{30} (2 x^{3} - 1)^{5}

Add a constant to the solution

= \frac{1}{30} (2 x^{3} - 1)^{5} + C

\int \frac{1}{x} + (x^{5} e^{x}) d x

\frac{\partial}{\partial x} (x^{4})

\frac{d}{d x} (- e^{- x} x - e^{- x})

\int 3 x^{2} (x^{3} + 7)^{3} d x

(9 x y^{2} - 3) d x + (9 x^{2} y + 4) d y = 0

What is the integral of x^2(2x^3-1)^4 ?
The integral of x^2(2x^3-1)^4 is 1/30 (2x^3-1)^5+C