What is the integral from 1 to 2 of (2^{ln(x)})/x ?

The integral from 1 to 2 of (2^{ln(x)})/x is (2^{ln(2)}-1)/(ln(2))

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integral from 1 to 2 of (2^{ln(x)})/x

\int_{1}^{2} \frac{2 ^{l n (x)}}{x} d x

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

Decimal Notation

0.88986 \dots

Solution steps

\int_{1}^{2} \frac{2 ^{l n (x)}}{x} d x

Apply u-substitution

: \int_{0}^{l n (2)} 2^{u} d u

= \int_{0}^{l n (2)} 2^{u} d u

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

= [\frac{2 ^{u}}{ln ( 2 )}]_{0}^{l n (2)}

Compute the boundaries:

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

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

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What is the integral from 1 to 2 of (2^{ln(x)})/x ?
The integral from 1 to 2 of (2^{ln(x)})/x is (2^{ln(2)}-1)/(ln(2))