What is the integral of 6x(4x^2+8)^{10} ?

The integral of 6x(4x^2+8)^{10} is 3/44 (4x^2+8)^{11}+C

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integral of 6x(4x^2+8)^{10}

\int 6 x (4 x^{2} + 8)^{10} d x

\frac{3}{44} (4 x^{2} + 8)^{11} + C

Solution steps

\int 6 x (4 x^{2} + 8)^{10} d x

Take the constant out:

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

= 6 \cdot \int x (4 x^{2} + 8)^{10} d x

Apply u-substitution

: \int \frac{u ^{10}}{8} d u

= 6 \cdot \int \frac{u ^{10}}{8} d u

Take the constant out:

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

= 6 \cdot \frac{1}{8} \cdot \int u^{10} d u

Apply the Power Rule

: \frac{u ^{11}}{11}

= 6 \cdot \frac{1}{8} \cdot \frac{u ^{11}}{11}

Substitute back

u = 4 x^{2} + 8

= 6 \cdot \frac{1}{8} \cdot \frac{( 4 x ^{2} + 8 ) ^{11}}{11}

Simplify

6 \cdot \frac{1}{8} \cdot \frac{( 4 x ^{2} + 8 ) ^{11}}{11} : \frac{3}{44} (4 x^{2} + 8)^{11}

= \frac{3}{44} (4 x^{2} + 8)^{11}

Add a constant to the solution

= \frac{3}{44} (4 x^{2} + 8)^{11} + C

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What is the integral of 6x(4x^2+8)^{10} ?
The integral of 6x(4x^2+8)^{10} is 3/44 (4x^2+8)^{11}+C