What is the integral of \sqrt[3]{4^{2x}} ?

The integral of \sqrt[3]{4^{2x}} is 1/(ln(2))((3*4^{(8x)/3})/(16^{x+1)}+9/16*4^{(2x)/3})+C

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integral of \sqrt[3]{4^{2x}}

\int 3 4^{2 x} d x

\frac{1}{ln ( 2 )} (\frac{3 \cdot 4 ^{\frac{8 x}{3}}}{1 6 ^{x + 1}} + \frac{9}{16} \cdot 4^{\frac{2 x}{3}}) + C

Solution steps

\int 3 4^{2 x} d x

Apply u-substitution

: \int \frac{3 u}{ln ( 2 ) \cdot 2 ^{\frac{l n ( u ) + 2 l n ( 2 )}{l n ( 2 )}}} d u

= \int \frac{3 u}{ln ( 2 ) \cdot 2 ^{\frac{l n ( u ) + 2 l n ( 2 )}{l n ( 2 )}}} d u

Take the constant out:

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

= \frac{1}{ln ( 2 )} \cdot \int \frac{3 u}{2 ^{\frac{l n ( u ) + 2 l n ( 2 )}{l n ( 2 )}}} d u

Apply Integration By Parts

: \frac{3 u ^{\frac{4}{3}}}{2 ^{\frac{l n ( u ) + 4 l n ( 2 )}{l n ( 2 )}}} - \int - \frac{3}{16 u ^{\frac{2}{3}}} d u

= \frac{1}{ln ( 2 )} (\frac{3 u ^{\frac{4}{3}}}{2 ^{\frac{l n ( u ) + 4 l n ( 2 )}{l n ( 2 )}}} - \int - \frac{3}{16 u ^{\frac{2}{3}}} d u)

\int - \frac{3}{16 u ^{\frac{2}{3}}} d u = - \frac{9}{16} u^{\frac{1}{3}}

= \frac{1}{ln ( 2 )} (\frac{3 u ^{\frac{4}{3}}}{2 ^{\frac{l n ( u ) + 4 l n ( 2 )}{l n ( 2 )}}} - (- \frac{9}{16} u^{\frac{1}{3}}))

Substitute back

u = 4^{2 x}

= \frac{1}{ln ( 2 )} \frac{3 ( 4 ^{2 x} ) ^{\frac{4}{3}}}{2 ^{\frac{l n ( 4 ^{2 x} ) + 4 l n ( 2 )}{l n ( 2 )}}} - (- \frac{9}{16} (4^{2 x})^{\frac{1}{3}})

Simplify

\frac{1}{ln ( 2 )} \frac{3 ( 4 ^{2 x} ) ^{\frac{4}{3}}}{2 ^{\frac{l n ( 4 ^{2 x} ) + 4 l n ( 2 )}{l n ( 2 )}}} - (- \frac{9}{16} (4^{2 x})^{\frac{1}{3}}) : \frac{1}{ln ( 2 )} (\frac{3 \cdot 4 ^{\frac{8 x}{3}}}{1 6 ^{x + 1}} + \frac{9}{16} \cdot 4^{\frac{2 x}{3}})

= \frac{1}{ln ( 2 )} (\frac{3 \cdot 4 ^{\frac{8 x}{3}}}{1 6 ^{x + 1}} + \frac{9}{16} \cdot 4^{\frac{2 x}{3}})

Add a constant to the solution

= \frac{1}{ln ( 2 )} (\frac{3 \cdot 4 ^{\frac{8 x}{3}}}{1 6 ^{x + 1}} + \frac{9}{16} \cdot 4^{\frac{2 x}{3}}) + C

3 e^{x} - 7, at x = 0

\frac{\partial}{\partial x} (\frac{- 8 x}{x ^{2} + 1})

y = 4 + 5 x^{2} - 2 x^{3}, at x = a

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

y^{^{'}} = 5 y + y^{2}

What is the integral of \sqrt[3]{4^{2x}} ?
The integral of \sqrt[3]{4^{2x}} is 1/(ln(2))((3*4^{(8x)/3})/(16^{x+1)}+9/16*4^{(2x)/3})+C