I'm having a problem trying to figure out how to integrate the following integral: ln(y)/y with respect to dy with limits of e and 1.
Please can someone assist me
Thanks
let $\displaystyle u = \ln y $
then $\displaystyle du = \frac{dy}{y} $
so the integral becomes $\displaystyle \int u \ du = \frac{u^{2}}{2} = \frac{(\ln y)^{2}}{2} \Big|^{e}_{1} $
$\displaystyle = \frac{1}{2} \Big((\ln e)^{2} - (\ln 1)^{2}\Big) = \frac{1}{2} \Big(1-0\Big) = \frac{1}{2}$
$\displaystyle \int{\frac{\ln{y}}{y}\,dy} = \int{\ln{y}\left(\frac{1}{y}\right)\,dy}$.
Now make the substitution $\displaystyle u = \ln{y}$ so that $\displaystyle \frac{du}{dy} = \frac{1}{y}$, the integral becomes
$\displaystyle \int{u\,\frac{du}{dy}\,dy}$
$\displaystyle = \int{u\,du}$
$\displaystyle = \frac{1}{2}u^2 + C$
$\displaystyle = \frac{1}{2}(\ln{y})^2 + C$.
Therefore:
$\displaystyle \int_1^e{\frac{\ln{y}}{y}\,dy}= \left[\frac{1}{2}(\ln{y})^2\right]_1^e$
$\displaystyle = \left[\frac{1}{2}(\ln{e})^2\right] - \left[\frac{1}{2}(\ln{1})^2\right]$
$\displaystyle = \frac{1}{2}$.