What is the integral of (2^{ln(z^3)})/(16z) ?

The integral of (2^{ln(z^3)})/(16z) is (2^{3ln(z)-4})/(3ln(2))+C

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integral of (2^{ln(z^3)})/(16z)

\int \frac{2 ^{l n (z^{3})}}{16 z} d z

\frac{2 ^{3 l n (z) - 4}}{3 ln ( 2 )} + C

Solution steps

\int \frac{2 ^{l n (z^{3})}}{16 z} d z

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{2 ^{l n (z^{3})}}{z} d z

Apply u-substitution

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

= \frac{1}{16} \cdot \int \frac{2 ^{l n (u)}}{3 u} d u

Take the constant out:

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

= \frac{1}{16} \cdot \frac{1}{3} \cdot \int \frac{2 ^{l n (u)}}{u} d u

Apply u-substitution

: \int 2^{v} d v

= \frac{1}{16} \cdot \frac{1}{3} \cdot \int 2^{v} d v

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

= \frac{1}{16} \cdot \frac{1}{3} \cdot \frac{2 ^{v}}{ln ( 2 )}

Substitute back

= \frac{1}{16} \cdot \frac{1}{3} \cdot \frac{2 ^{l n (z^{3})}}{ln ( 2 )}

Simplify

\frac{1}{16} \cdot \frac{1}{3} \cdot \frac{2 ^{l n (z^{3})}}{ln ( 2 )} : \frac{2 ^{3 l n (z) - 4}}{3 ln ( 2 )}

= \frac{2 ^{3 l n (z) - 4}}{3 ln ( 2 )}

Add a constant to the solution

= \frac{2 ^{3 l n (z) - 4}}{3 ln ( 2 )} + C

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What is the integral of (2^{ln(z^3)})/(16z) ?
The integral of (2^{ln(z^3)})/(16z) is (2^{3ln(z)-4})/(3ln(2))+C