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

The integral of 3x^2(2x^3+4)^7 is 1/16 (2x^3+4)^8+C

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integral of 3x^2(2x^3+4)^7

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

\frac{1}{16} (2 x^{3} + 4)^{8} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{7}}{6} d u

= 3 \cdot \int \frac{u ^{7}}{6} d u

Take the constant out:

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

= 3 \cdot \frac{1}{6} \cdot \int u^{7} d u

Apply the Power Rule

: \frac{u ^{8}}{8}

= 3 \cdot \frac{1}{6} \cdot \frac{u ^{8}}{8}

Substitute back

u = 2 x^{3} + 4

= 3 \cdot \frac{1}{6} \cdot \frac{( 2 x ^{3} + 4 ) ^{8}}{8}

Simplify

3 \cdot \frac{1}{6} \cdot \frac{( 2 x ^{3} + 4 ) ^{8}}{8} : \frac{1}{16} (2 x^{3} + 4)^{8}

= \frac{1}{16} (2 x^{3} + 4)^{8}

Add a constant to the solution

= \frac{1}{16} (2 x^{3} + 4)^{8} + C

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What is the integral of 3x^2(2x^3+4)^7 ?
The integral of 3x^2(2x^3+4)^7 is 1/16 (2x^3+4)^8+C