What is the integral of 21x^2(7x^3-8)^6 ?

The integral of 21x^2(7x^3-8)^6 is 1/7 (7x^3-8)^7+C

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integral of 21x^2(7x^3-8)^6

\int 21 x^{2} (7 x^{3} - 8)^{6} d x

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

Solution steps

\int 21 x^{2} (7 x^{3} - 8)^{6} d x

Take the constant out:

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

= 21 \cdot \int x^{2} (7 x^{3} - 8)^{6} d x

Apply u-substitution

: \int \frac{u ( 7 u ^{\frac{3}{2}} - 8 ) ^{6}}{2 x} d u

= 21 \cdot \int \frac{( 7 u - 8 ) ^{6}}{3} d u

Take the constant out:

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

= 21 \cdot \frac{1}{3} \cdot \int (7 u - 8)^{6} d u

Apply u-substitution

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

= 21 \cdot \frac{1}{3} \cdot \int \frac{v ^{6}}{7} d v

Take the constant out:

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

= 21 \cdot \frac{1}{3} \cdot \frac{1}{7} \cdot \int v^{6} d v

Apply the Power Rule

: \frac{v ^{7}}{7}

= 21 \cdot \frac{1}{3} \cdot \frac{1}{7} \cdot \frac{v ^{7}}{7}

Substitute back

= 21 \cdot \frac{1}{3} \cdot \frac{1}{7} \cdot \frac{( 7 x ^{3} - 8 ) ^{7}}{7}

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 21x^2(7x^3-8)^6 ?
The integral of 21x^2(7x^3-8)^6 is 1/7 (7x^3-8)^7+C