What is the integral of 3x^2(x^3+7)^7 ?

The integral of 3x^2(x^3+7)^7 is 1/8 (x^3+7)^8+C

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integral of 3x^2(x^3+7)^7

\int 3 x^{2} (x^{3} + 7)^{7} d x

\frac{1}{8} (x^{3} + 7)^{8} + C

Solution steps

\int 3 x^{2} (x^{3} + 7)^{7} 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} + 7)^{7} d x

Apply u-substitution

= 3 \cdot \int \frac{u ^{7}}{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^{7} d u

Apply the Power Rule

= 3 \cdot \frac{1}{3} \cdot \frac{u ^{8}}{8}

Substitute back

u = x^{3} + 7

= 3 \cdot \frac{1}{3} \cdot \frac{( x ^{3} + 7 ) ^{8}}{8}

Simplify

3 \cdot \frac{1}{3} \cdot \frac{( x ^{3} + 7 ) ^{8}}{8} : \frac{1}{8} (x^{3} + 7)^{8}

= \frac{1}{8} (x^{3} + 7)^{8}

Add a constant to the solution

= \frac{1}{8} (x^{3} + 7)^{8} + C

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What is the integral of 3x^2(x^3+7)^7 ?
The integral of 3x^2(x^3+7)^7 is 1/8 (x^3+7)^8+C