What is the integral of 15x^2(x^3-1)^4 ?

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

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integral of 15x^2(x^3-1)^4

\int 15 x^{2} (x^{3} - 1)^{4} d x

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

Solution steps

\int 15 x^{2} (x^{3} - 1)^{4} d x

Take the constant out:

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

= 15 \cdot \int x^{2} (x^{3} - 1)^{4} d x

Apply u-substitution

: \int \frac{u ^{4}}{3} d u

= 15 \cdot \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

= 15 \cdot \frac{1}{3} \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= 15 \cdot \frac{1}{3} \cdot \frac{u ^{5}}{5}

Substitute back

u = x^{3} - 1

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

Simplify

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

= (x^{3} - 1)^{5}

Add a constant to the solution

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

- 15 x^{4} + 270 x^{2}

\frac{d}{d x} (5 e^{6 x})

\int 8 x^{3} e^{- x^{2}} d x

y^{^{'}} = y (x y^{4} + 2)

x \to 9 lim ((x - 9)^{4})

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