What is the integral of 15x^2(x^3-7)^4 ?

The integral of 15x^2(x^3-7)^4 is (x^3-7)^5+C

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integral of 15x^2(x^3-7)^4

\int 15 x^{2} (x^{3} - 7)^{4} d x

(x^{3} - 7)^{5} + C

Solution steps

\int 15 x^{2} (x^{3} - 7)^{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} - 7)^{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} - 7

= 15 \cdot \frac{1}{3} \cdot \frac{( x ^{3} - 7 ) ^{5}}{5}

Simplify

15 \cdot \frac{1}{3} \cdot \frac{( x ^{3} - 7 ) ^{5}}{5} : (x^{3} - 7)^{5}

= (x^{3} - 7)^{5}

Add a constant to the solution

= (x^{3} - 7)^{5} + C

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What is the integral of 15x^2(x^3-7)^4 ?
The integral of 15x^2(x^3-7)^4 is (x^3-7)^5+C