What is the integral of 2^{ln(x)} ?

The integral of 2^{ln(x)} is ((2e)^{ln(x)})/(ln(2e))+C

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integral of 2^{ln(x)}

\int 2^{l n (x)} d x

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

Solution steps

\int 2^{l n (x)} d x

Apply u-substitution

: \int 2^{u} e^{u} d u

= \int 2^{u} e^{u} d u

Apply exponent rule:

a^{n} \cdot b^{n} = (a \cdot b)^{n}

2^{u} e^{u} = (2 e)^{u}

= \int (2 e)^{u} d u

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

= \frac{( 2 e ) ^{u}}{ln ( 2 e )}

Substitute back

u = ln (x)

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

Add a constant to the solution

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

f (x) = x - 2

\int_{0}^{L} sin^{2} (\frac{π x}{L}) d x

\frac{d}{d x} ((\frac{3}{5} x)^{3})

f (x) = 2 x + 1

n = 0 \sum \infty \frac{1}{n n}

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