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

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

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integral of (4^{ln(1/x)})/x

\int \frac{4 ^{l n (\frac{1}{x})}}{x} d x

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

Solution steps

\int \frac{4 ^{l n (\frac{1}{x})}}{x} d x

Apply u-substitution

: \int - \frac{4 ^{l n (u)}}{u} d u

= \int - \frac{4 ^{l n (u)}}{u} d u

Take the constant out:

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

= - \int \frac{4 ^{l n (u)}}{u} d u

Apply u-substitution

: \int 4^{v} d v

= - \int 4^{v} d v

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

= - \frac{4 ^{v}}{ln ( 4 )}

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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