What is the integral of (4x^3+7)^{12}x^2 ?

The integral of (4x^3+7)^{12}x^2 is 1/156 (4x^3+7)^{13}+C

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integral of (4x^3+7)^{12}x^2

\int (4 x^{3} + 7)^{12} x^{2} d x

\frac{1}{156} (4 x^{3} + 7)^{13} + C

Solution steps

\int (4 x^{3} + 7)^{12} x^{2} d x

Apply u-substitution

: \int \frac{( 4 u + 7 ) ^{12}}{3} d u

= \int \frac{( 4 u + 7 ) ^{12}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int (4 u + 7)^{12} d u

Apply u-substitution

: \int \frac{v ^{12}}{4} d v

= \frac{1}{3} \cdot \int \frac{v ^{12}}{4} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{4} \cdot \int v^{12} d v

Apply the Power Rule

: \frac{v ^{13}}{13}

= \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{v ^{13}}{13}

Substitute back

= \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{( 4 x ^{3} + 7 ) ^{13}}{13}

Simplify

\frac{1}{3} \cdot \frac{1}{4} \cdot \frac{( 4 x ^{3} + 7 ) ^{13}}{13} : \frac{1}{156} (4 x^{3} + 7)^{13}

= \frac{1}{156} (4 x^{3} + 7)^{13}

Add a constant to the solution

= \frac{1}{156} (4 x^{3} + 7)^{13} + C

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What is the integral of (4x^3+7)^{12}x^2 ?
The integral of (4x^3+7)^{12}x^2 is 1/156 (4x^3+7)^{13}+C