What is the integral of x^2*6^{x^3} ?

The integral of x^2*6^{x^3} is (2^{x^3}*3^{x^3-1})/(ln(6))+C

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integral of x^2*6^{x^3}

\int x^{2} \cdot 6^{x^{3}} d x

\frac{2 ^{x^{3}} \cdot 3 ^{x^{3} - 1}}{ln ( 6 )} + C

Solution steps

\int x^{2} \cdot 6^{x^{3}} d x

Apply u-substitution

: \int \frac{3 ^{u^{\frac{3}{2}}} \cdot 2 ^{u^{\frac{3}{2}} - 1} u}{x} d u

= \int 2^{u} \cdot 3^{u - 1} d u

Simplify

3^{u - 1} : 3^{u} \cdot 3^{- 1}

= \int 2^{u} \cdot 3^{u} \cdot 3^{- 1} d u

Take the constant out:

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

= 3^{- 1} \cdot \int 2^{u} \cdot 3^{u} d u

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{3} \cdot \int 2^{u} \cdot 3^{u} d u

Simplify

2^{u} \cdot 3^{u} : 6^{u}

= \frac{1}{3} \cdot \int 6^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{1}{3} \cdot \frac{6 ^{u}}{ln ( 6 )}

Substitute back

u = x^{3}

= \frac{1}{3} \cdot \frac{6 ^{x^{3}}}{ln ( 6 )}

Simplify

\frac{1}{3} \cdot \frac{6 ^{x^{3}}}{ln ( 6 )} : \frac{2 ^{x^{3}} \cdot 3 ^{x^{3} - 1}}{ln ( 6 )}

= \frac{2 ^{x^{3}} \cdot 3 ^{x^{3} - 1}}{ln ( 6 )}

Add a constant to the solution

= \frac{2 ^{x^{3}} \cdot 3 ^{x^{3} - 1}}{ln ( 6 )} + C

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What is the integral of x^2*6^{x^3} ?
The integral of x^2*6^{x^3} is (2^{x^3}*3^{x^3-1})/(ln(6))+C