What is the integral of 6(x^{sqrt(3)})^{+3} ?

The integral of 6(x^{sqrt(3)})^{+3} is (3(3sqrt(3)-1))/(13)x^{3sqrt(3)+1}+C

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integral of 6(x^{sqrt(3)})^{+3}

\int 6 (x^{3})^{+ 3} d x

\frac{3 ( 3 3 - 1 )}{13} x^{33 + 1} + C

Solution steps

\int 6 (x^{3})^{3} d x

Take the constant out:

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

= 6 \cdot \int (x^{3})^{3} d x

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 6 \cdot \int x^{33} d x

Apply the Power Rule

: \frac{3 3 - 1}{26} x^{33 + 1}

= 6 \cdot \frac{3 3 - 1}{26} x^{33 + 1}

Simplify

6 \cdot \frac{3 3 - 1}{26} x^{33 + 1} : \frac{3 ( 3 3 - 1 )}{13} x^{33 + 1}

= \frac{3 ( 3 3 - 1 )}{13} x^{33 + 1}

Add a constant to the solution

= \frac{3 ( 3 3 - 1 )}{13} x^{33 + 1} + C

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What is the integral of 6(x^{sqrt(3)})^{+3} ?
The integral of 6(x^{sqrt(3)})^{+3} is (3(3sqrt(3)-1))/(13)x^{3sqrt(3)+1}+C