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

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

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

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

\frac{1}{15} (x^{3} - 2)^{5} + C

Solution steps

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

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

Simplify

\frac{1}{3} \cdot \frac{( x ^{3} - 2 ) ^{5}}{5} : \frac{1}{15} (x^{3} - 2)^{5}

= \frac{1}{15} (x^{3} - 2)^{5}

Add a constant to the solution

= \frac{1}{15} (x^{3} - 2)^{5} + C

\int \frac{( ln ( x ^{5} ) )}{x} d x

\frac{\partial}{\partial y} (7 x^{2} e^{y})

f (x) = x^{3} x^{4} + 3 x^{2} - 2

e^{x}, e^{- x}, 0, 1

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

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