# Integrals

#### 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

#### Random Variable

let $$\displaystyle u = \ln y$$

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

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

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

#### Prove It

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

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:

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

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

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

Thank you