That is not right: $\displaystyle du=-dx/x^2$. Better use the substitution $\displaystyle t=\ln x$ . Then:
$\displaystyle \displaystyle\int_0^e 2^{\ln x}\cdot \dfrac{dx}{x}=\displaystyle\int_0^12^tdt=\ldots $
Could you continue?.
Regards.
Fernando Revilla
because ln(x) is the integral of 1/x, not the derivative. If u= 1/x then $\displaystyle du= -1/x^2 dx$ not "ln x dx". That is precisely what FernandoRevilla said before. Instead of integration by parts, use the simple substitution t= ln x that FernandoRevilla suggested.
$\displaystyle \displaystyle\int_0^1 2^tdt=\left[\dfrac{2^t}{\ln 2}\right]_0^1=\dfrac{1}{\ln 2}$
Regards.
Fernando Revilla
You are collectivly making a meal of this:
$\displaystyle \displaystyle \int_1^e \frac{2^{\ln(x)}}{x}\;dx=\int_1^e x^{\ln(2)-1}\;dx=\left[ \frac{1}{\ln(2)} x^{\ln(2)}\right]_1^e$
............. $\displaystyle \displaystyle =\frac{1}{\ln(2)}(e^{\ln(2)}-1^{\ln(2)})=\frac{1}{\ln(2)}(2-1)=\frac{1}{\ln(2)}$
CB