# How to integrate int 2^lnx/xdx ?

Jun 23, 2018

${2}^{\log} \frac{x}{\log} \left(2\right) + C$

#### Explanation:

Substituting $\log \left(x\right) = t$ then

$\frac{1}{x} \mathrm{dx} = \mathrm{dt}$
so we get

$\int {2}^{t} \mathrm{dt} = {2}^{t} / \log \left(2\right) + C$

Aug 9, 2018

$\int \frac{{2}^{\ln} x}{x} \mathrm{dx} = \frac{{2}^{\ln} x}{\ln 2} + C$

#### Explanation:

$\int \frac{{2}^{\ln} x}{x} \mathrm{dx} = \int \left({2}^{\ln x}\right) \left(\frac{1}{x}\right) \mathrm{dx}$

Use a u-substitution:

$\textcolor{b l u e}{u = \ln x}$

$\textcolor{b l u e}{\mathrm{du} = \frac{1}{x} \mathrm{dx}}$

$\int \left({2}^{\textcolor{b l u e}{u}}\right) \mathrm{du}$

$= \frac{{2}^{u}}{\ln 2} + C$

Substitute $\textcolor{b l u e}{u = \ln x}$ back in:

$\int \frac{{2}^{\ln} x}{x} \mathrm{dx} = \frac{{2}^{\ln} x}{\ln 2} + C$

Aug 9, 2018

${2}^{\ln} \frac{x}{\ln} 2 + C$

#### Explanation:

U-sub where $u = \ln \left(x\right) \mathrm{dx}$

$u = \ln x \mathrm{dx}$

$\mathrm{du} = \frac{1}{x} \mathrm{dx}$

$\int {2}^{u} \mathrm{du}$
$=$
${2}^{u} / \ln 2 + C$

sub $\ln \left(x\right)$ back in

${2}^{\ln} \frac{x}{\ln} 2 + C$