Definite Integral..Clarify my work.. My professor worked this solution in class, but I am not sure he is right. I did it on my own if someone could take a look at my work. I am a little confused.
$$\int_1^2 \frac{\ln(x)}{x}  \;dx$$
So I approached this problem with $U$ Substitution. So
$u= \ln(x)$   which means $\;du = \frac{1}{x} \;dx$
Next
I find the new bounds for the integral since I used $U$ Substitution. So,
instead of the lower limit being $1$, it would be $\ln(1)$, and instead of the upper limit being $2$ it would be $\ln(2)$.
$$1 = \ln(1)$$
$$2 = \ln(2)$$
For new definite integral limits, I simplified the integral to $u \;du$ since I substituted in $\;du = \frac{1}{x} \;dx$, integrating $u \;du$ I get 
$$\frac{u^2}{2}.$$
Which substituting my original $u$ back in I get $$\frac{\ln(x)^2}{2}.$$
So my big question is: now that I have this equation, to do $f(b) - f(a)$ would I just plug in the lower and upper bounds so it would look something like
$$\frac{\ln(\ln(2))^2 }{2} - \frac{\ln(\ln(1))^2}{2}$$
 A: Once you perform a substitution on a definite integral, you change the limits accordingly and then you don't need to go back to the original limits. In your case, $u = \ln(x)$ and
$$ \int_1^2 \frac{\ln(x)}{x} \, dx = \int_{\ln(1)}^{\ln(2)} u \, du = \left[ \frac{u^2}{2} \right]_{u = \ln(1)}^{u = \ln(2)} = \frac{\ln^2(2)}{2} $$
(where we used $\ln(1) = 0$).
If you were to find the indefinite integral of $\frac{\ln(x)}{x}$, then you would have to return to the original variable in the end:
$$ \int \frac{\ln(x)}{x} \, dx = \int u \, du = \frac{u^2}{2} + C = \frac{\ln^2(x)}{2} + C. $$
A primitive of $\frac{\ln(x)}{x}$ on $(0,\infty)$ is given by $\frac{\ln^2(x)}{2}$ and so we also have
$$ \int_1^2 \frac{\ln(x)}{x} \, dx = \left[ \frac{\ln^2(x)}{2} \right]_{x = 1}^{x = 2} = \frac{\ln^2(2)}{2} $$
which is consistent with what we have done before.
A: Let $$I=\int\limits_1^2 \frac{\ln x}{x}dx $$
now put $u=\ln x$. This means $du=\frac{dx}{x}$ and $dx=e^{u}du$. The integral becomes
\begin{align}
\int\limits_{\ln1}^{\ln 2}u du &= \frac{u^{2}}{2}\Biggr\rvert_{\ln 1}^{\ln 2} \\
                               &= \frac{1}{2}\left(\ln^{2}2-\ln^{2}1 \right) \\
                               &= \frac{1}{2}\ln^{2}2
\end{align}
