What is the integral of (3-4x)(4x^2-6x+7)^{10} ?

The integral of (3-4x)(4x^2-6x+7)^{10} is -1/92274688 (16x^2-24x+28)^{11}+C

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integral of (3-4x)(4x^2-6x+7)^{10}

\int (3 - 4 x) (4 x^{2} - 6 x + 7)^{10} d x

- \frac{1}{92274688} (16 x^{2} - 24 x + 28)^{11} + C

Solution steps

\int (3 - 4 x) (4 x^{2} - 6 x + 7)^{10} d x

Apply u-substitution

: \int - \frac{u ( u ^{2} + 19 ) ^{10}}{4194304} d u

= \int - \frac{u ( u ^{2} + 19 ) ^{10}}{4194304} d u

Take the constant out:

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

= - \frac{1}{4194304} \cdot \int u (u^{2} + 19)^{10} d u

Apply u-substitution

: \int \frac{v ^{10}}{2} d v

= - \frac{1}{4194304} \cdot \int \frac{v ^{10}}{2} d v

Take the constant out:

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

= - \frac{1}{4194304} \cdot \frac{1}{2} \cdot \int v^{10} d v

Apply the Power Rule

: \frac{v ^{11}}{11}

= - \frac{1}{4194304} \cdot \frac{1}{2} \cdot \frac{v ^{11}}{11}

Substitute back

= - \frac{1}{4194304} \cdot \frac{1}{2} \cdot \frac{( ( 3 - 4 x ) ^{2} + 19 ) ^{11}}{11}

Simplify

- \frac{1}{4194304} \cdot \frac{1}{2} \cdot \frac{( ( 3 - 4 x ) ^{2} + 19 ) ^{11}}{11} : - \frac{1}{92274688} (16 x^{2} - 24 x + 28)^{11}

= - \frac{1}{92274688} (16 x^{2} - 24 x + 28)^{11}

Add a constant to the solution

= - \frac{1}{92274688} (16 x^{2} - 24 x + 28)^{11} + C

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What is the integral of (3-4x)(4x^2-6x+7)^{10} ?
The integral of (3-4x)(4x^2-6x+7)^{10} is -1/92274688 (16x^2-24x+28)^{11}+C