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

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

Popular >

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\int x e^{- 8 x^{2}} d x

\int 3^{- x} cos (x) d x

\int 1 + \frac{1}{x} d x

\frac{\partial}{\partial y} (5 x^{2} + 8 x y + y^{2})

\frac{d y}{d x} + \frac{5 x ^{2}}{e ^{3 x^{3}}} y = 0

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