What is the integral of 1/(4^x) ?

The integral of 1/(4^x) is -1/(ln(2)*2^{2x+1)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(4^x)

\int \frac{1}{4 ^{x}} d x

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

Solution steps

\int \frac{1}{4 ^{x}} d x

Apply u-substitution

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

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

\frac{1}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}} u} = \frac{1 - 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}} u}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}} u} + 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d u = u

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

Substitute back

u = 4^{x}

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

Simplify

\frac{1}{ln ( 2 )} \frac{- 1 - 2 ^{\frac{2 l n ( 4 ^{x} ) + l n ( 2 )}{l n ( 2 )}}}{2 ^{\frac{l n ( 4 ^{x} ) + l n ( 2 )}{l n ( 2 )}}} + 4^{x} : - \frac{1}{ln ( 2 ) \cdot 2 ^{2 x + 1}}

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

Add a constant to the solution

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

\int x e^{- 6 x^{2}} d x

x \to - 2 lim (4 x^{2} - 3 x + 5)

\frac{y ^{2} + 1}{y d y} = \frac{x}{d x}

\frac{d}{d x} (2 x^{2} - 4 x + 5)

\frac{d}{d x} (e^{x} (2 x^{4} + 3 cos (x)))

What is the integral of 1/(4^x) ?
The integral of 1/(4^x) is -1/(ln(2)*2^{2x+1)}+C