What is the integral of \sqrt[5]{2^{3x}} ?

The integral of \sqrt[5]{2^{3x}} is (5*8^{x/5})/(3ln(2))+C

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integral of \sqrt[5]{2^{3x}}

\frac{5 \cdot 8 ^{\frac{x}{5}}}{3 ln ( 2 )} + C

Solution steps

Apply exponent rule:

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

\frac{1}{5} = 5^{- 1}

= \int (2^{3 x})^{5^{- 1}} d x

Simplify

(2^{3 x})^{5^{- 1}} : 8^{\frac{x}{5}}

= \int 8^{\frac{x}{5}} d x

Apply u-substitution

= \int 8^{u} \cdot 5 d u

Take the constant out:

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

= 5 \cdot \int 8^{u} d u

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

= 5 \cdot \frac{8 ^{u}}{ln ( 8 )}

Substitute back

u = \frac{x}{5}

= 5 \cdot \frac{8 ^{\frac{x}{5}}}{ln ( 8 )}

Simplify

5 \cdot \frac{8 ^{\frac{x}{5}}}{ln ( 8 )} : \frac{5 \cdot 8 ^{\frac{x}{5}}}{3 ln ( 2 )}

= \frac{5 \cdot 8 ^{\frac{x}{5}}}{3 ln ( 2 )}

Add a constant to the solution

= \frac{5 \cdot 8 ^{\frac{x}{5}}}{3 ln ( 2 )} + C

What is the integral of \sqrt[5]{2^{3x}} ?
The integral of \sqrt[5]{2^{3x}} is (5*8^{x/5})/(3ln(2))+C