What is the integral of 6x^3(-3x^4-7)^6 ?

The integral of 6x^3(-3x^4-7)^6 is 1/14 (3x^4+7)^7+C

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integral of 6x^3(-3x^4-7)^6

\int 6 x^{3} (- 3 x^{4} - 7)^{6} d x

\frac{1}{14} (3 x^{4} + 7)^{7} + C

Solution steps

\int 6 x^{3} (- 3 x^{4} - 7)^{6} d x

Take the constant out:

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

= 6 \cdot \int x^{3} (- 3 x^{4} - 7)^{6} d x

Factor

- 3 x^{4} - 7 : - (3 x^{4} + 7)

= 6 \cdot \int x^{3} (- (3 x^{4} + 7))^{6} d x

(- (3 x^{4} + 7))^{6} = (3 x^{4} + 7)^{6}

= 6 \cdot \int x^{3} (3 x^{4} + 7)^{6} d x

Apply u-substitution

: \int \frac{u ^{6}}{12} d u

= 6 \cdot \int \frac{u ^{6}}{12} d u

Take the constant out:

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

= 6 \cdot \frac{1}{12} \cdot \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= 6 \cdot \frac{1}{12} \cdot \frac{u ^{7}}{7}

Substitute back

u = 3 x^{4} + 7

= 6 \cdot \frac{1}{12} \cdot \frac{( 3 x ^{4} + 7 ) ^{7}}{7}

Simplify

6 \cdot \frac{1}{12} \cdot \frac{( 3 x ^{4} + 7 ) ^{7}}{7} : \frac{1}{14} (3 x^{4} + 7)^{7}

= \frac{1}{14} (3 x^{4} + 7)^{7}

Add a constant to the solution

= \frac{1}{14} (3 x^{4} + 7)^{7} + C

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What is the integral of 6x^3(-3x^4-7)^6 ?
The integral of 6x^3(-3x^4-7)^6 is 1/14 (3x^4+7)^7+C