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

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

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

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

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

Solution steps

\int 3 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

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

Apply u-substitution

= 3 \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

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

Apply the Power Rule

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

Substitute back

u = x^{3} - 1

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

Simplify

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

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

Add a constant to the solution

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

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