TITLE: How to integrate $\frac{1}{2^{(\ln x)}}$?
QUESTION [7 upvotes]: 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.

REPLY [6 votes]: There is no substitution required. Notice that $2^{\ln(x)} = x^{\ln(2)}$ for all $x > 0$. Therefore,
$$
\int \dfrac{1}{2^{\ln(x)}} \,\mathrm{d}x = \int x^{-\ln(2)} \,\mathrm{d}x = \dfrac{x^{1-\ln(2)}}{1-\ln(2)} + c \qquad\text{(for a $c \in \mathbb{R}$).}
$$