What is the integral of 5x(6x^2-7)^4 ?

The integral of 5x(6x^2-7)^4 is 1/12 (6x^2-7)^5+C

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integral of 5x(6x^2-7)^4

\int 5 x (6 x^{2} - 7)^{4} d x

\frac{1}{12} (6 x^{2} - 7)^{5} + C

Solution steps

\int 5 x (6 x^{2} - 7)^{4} d x

Take the constant out:

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

= 5 \cdot \int x (6 x^{2} - 7)^{4} d x

Apply u-substitution

= 5 \cdot \int \frac{( 6 u - 7 ) ^{4}}{2} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int (6 u - 7)^{4} d u

Apply u-substitution

= 5 \cdot \frac{1}{2} \cdot \int \frac{v ^{4}}{6} d v

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \int v^{4} d v

Apply the Power Rule

= 5 \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \frac{v ^{5}}{5}

Substitute back

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

Simplify

5 \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \frac{( 6 x ^{2} - 7 ) ^{5}}{5} : \frac{1}{12} (6 x^{2} - 7)^{5}

= \frac{1}{12} (6 x^{2} - 7)^{5}

Add a constant to the solution

= \frac{1}{12} (6 x^{2} - 7)^{5} + C

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What is the integral of 5x(6x^2-7)^4 ?
The integral of 5x(6x^2-7)^4 is 1/12 (6x^2-7)^5+C