What is the integral of (x^3+8)^53x^2 ?

The integral of (x^3+8)^53x^2 is 1/6 (x^3+8)^6+C

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integral of (x^3+8)^53x^2

\int (x^{3} + 8)^{5} 3 x^{2} d x

\frac{1}{6} (x^{3} + 8)^{6} + C

Solution steps

\int (x^{3} + 8)^{5} \cdot 3 x^{2} d x

Take the constant out:

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

= 3 \cdot \int (x^{3} + 8)^{5} x^{2} d x

Apply u-substitution

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

Apply the Power Rule

= 3 \cdot \frac{1}{3} \cdot \frac{u ^{6}}{6}

Substitute back

u = x^{3} + 8

= 3 \cdot \frac{1}{3} \cdot \frac{( x ^{3} + 8 ) ^{6}}{6}

Simplify

3 \cdot \frac{1}{3} \cdot \frac{( x ^{3} + 8 ) ^{6}}{6} : \frac{1}{6} (x^{3} + 8)^{6}

= \frac{1}{6} (x^{3} + 8)^{6}

Add a constant to the solution

= \frac{1}{6} (x^{3} + 8)^{6} + C

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What is the integral of (x^3+8)^53x^2 ?
The integral of (x^3+8)^53x^2 is 1/6 (x^3+8)^6+C