How to integrate $\frac{1}{2^{(\ln x)}}$?

Question

How to integrate

$$\frac{1}{2^{(\ln x)}}$$?

I tried using substitution $u=\ln x$, but $du=\frac{dx}{x}$ is not in the original equation.

Answer 1

There is no substitution required. Notice that $2^{\ln(x)} = x^{\ln(2)}$ for all $x > 0$. Therefore, $$ \int \dfrac{1}{2^{\ln(x)}} \,\mathrm{d}x = \int x^{-\ln(2)} \,\mathrm{d}x = \dfrac{x^{1-\ln(2)}}{1-\ln(2)} + c \qquad\text{(for a $c \in \mathbb{R}$).} $$

Answer 2

One may observe that, by the change of variable $x=e^u$, $dx=e^u\:du$, $\ln x=u$, $$ \int\frac1{2^{\ln x}}\:dx=\int\frac1{2^{u}}\:e^u\:du=\int e^{(1-\ln 2)u}\:du $$ which is easier to evaluate.

Answer 3

Here's a different approach with some logarithm rules $$\int\frac{1}{2^{\ln x}}dx=\int\frac{1}{2^{\log_2 x/\log_2e}}dx=\int\frac{1}{(2^{\log_2 x})^{1/\log_2 e}}dx=\int \frac{1}{x^{1/\log_2 e}}dx=\frac{x^{1-1/\log_2 e}}{1-1/\log_2 e}=\frac{x^{1-\ln(2)}}{1-\ln(2)}$$