1. ## Integrals

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.

Thanks

2. let $u = \ln y$

then $du = \frac{dy}{y}$

so the integral becomes $\int u \ du = \frac{u^{2}}{2} = \frac{(\ln y)^{2}}{2} \Big|^{e}_{1}$

$= \frac{1}{2} \Big((\ln e)^{2} - (\ln 1)^{2}\Big) = \frac{1}{2} \Big(1-0\Big) = \frac{1}{2}$

3. Originally Posted by Nas
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.

Thanks
$\int{\frac{\ln{y}}{y}\,dy} = \int{\ln{y}\left(\frac{1}{y}\right)\,dy}$.

Now make the substitution $u = \ln{y}$ so that $\frac{du}{dy} = \frac{1}{y}$, the integral becomes

$\int{u\,\frac{du}{dy}\,dy}$

$= \int{u\,du}$

$= \frac{1}{2}u^2 + C$

$= \frac{1}{2}(\ln{y})^2 + C$.

Therefore:

$\int_1^e{\frac{\ln{y}}{y}\,dy}= \left[\frac{1}{2}(\ln{y})^2\right]_1^e$

$= \left[\frac{1}{2}(\ln{e})^2\right] - \left[\frac{1}{2}(\ln{1})^2\right]$

$= \frac{1}{2}$.

4. Thank you