What is the integral of \sqrt[5]{4^{3x}} ?

The integral of \sqrt[5]{4^{3x}} is (5*2^{1/5 (6x-5)})/(3ln(2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of \sqrt[5]{4^{3x}}

\frac{5 \cdot 2 ^{\frac{1}{5} (6 x - 5)}}{3 ln ( 2 )} + C

Solution steps

Apply radical rule:

assuming

a \geq 0

= \int 4^{\frac{3 x}{5}} d x

Apply u-substitution

: \int 4^{u} \cdot \frac{5}{3} d u

= \int 4^{u} \cdot \frac{5}{3} d u

Take the constant out:

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

= \frac{5}{3} \cdot \int 4^{u} d u

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

= \frac{5}{3} \cdot \frac{4 ^{u}}{ln ( 4 )}

Substitute back

u = \frac{3 x}{5}

= \frac{5}{3} \cdot \frac{4 ^{\frac{3 x}{5}}}{ln ( 4 )}

Simplify

\frac{5}{3} \cdot \frac{4 ^{\frac{3 x}{5}}}{ln ( 4 )} : \frac{5 \cdot 2 ^{\frac{1}{5} (6 x - 5)}}{3 ln ( 2 )}

= \frac{5 \cdot 2 ^{\frac{1}{5} (6 x - 5)}}{3 ln ( 2 )}

Add a constant to the solution

= \frac{5 \cdot 2 ^{\frac{1}{5} (6 x - 5)}}{3 ln ( 2 )} + C

y = 9 - 0.25 x^{2}, y = 1, 2 < x < 5

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + 9 y ^{2}})

\int (x + 1 + x^{2})^{6} d x

\int sec^{4} (5 x) tan (5 x) d x

x \to 2 lim (sin (\frac{p i x}{2}))

What is the integral of \sqrt[5]{4^{3x}} ?
The integral of \sqrt[5]{4^{3x}} is (5*2^{1/5 (6x-5)})/(3ln(2))+C