What is the integral of 5/2 x^2(x^3-7)^4 ?

The integral of 5/2 x^2(x^3-7)^4 is 1/6 (x^3-7)^5+C

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integral of 5/2 x^2(x^3-7)^4

\int \frac{5}{2} x^{2} (x^{3} - 7)^{4} d x

\frac{1}{6} (x^{3} - 7)^{5} + C

Solution steps

\int \frac{5}{2} 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

= \frac{5}{2} \cdot \int x^{2} (x^{3} - 7)^{4} d x

Apply u-substitution

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

= \frac{5}{2} \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

= \frac{5}{2} \cdot \frac{1}{3} \cdot \int u^{4} d u

Apply the Power Rule

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

= \frac{5}{2} \cdot \frac{1}{3} \cdot \frac{u ^{5}}{5}

Substitute back

u = x^{3} - 7

= \frac{5}{2} \cdot \frac{1}{3} \cdot \frac{( x ^{3} - 7 ) ^{5}}{5}

Simplify

\frac{5}{2} \cdot \frac{1}{3} \cdot \frac{( x ^{3} - 7 ) ^{5}}{5} : \frac{1}{6} (x^{3} - 7)^{5}

= \frac{1}{6} (x^{3} - 7)^{5}

Add a constant to the solution

= \frac{1}{6} (x^{3} - 7)^{5} + C

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\int x^{3} 5 - x^{4} d x

What is the integral of 5/2 x^2(x^3-7)^4 ?
The integral of 5/2 x^2(x^3-7)^4 is 1/6 (x^3-7)^5+C