Answer
$a.\quad\sinh^{-1}1$
$b.\quad\ln(1+\sqrt{2})$
Work Step by Step
Substitute:$\quad \left[\begin{array}{ll}
u=\ln xx & du=\frac{1}{x}dx\\
x=1 & \rightarrow u=0\\
x=e & \rightarrow u=1
\end{array}\right]$
$I=\displaystyle \int_{1}^{e}\frac{dx}{x\sqrt{1+(\ln x)^{2}}}=\int_{0}^{1}\frac{du}{\sqrt{1+u^{2}}}$
Using the table "lntegrals leading to inverse hyperbolic functions"
$\displaystyle \int\frac{du}{\sqrt{a^{2}+u^{2}}}=\sinh^{-1}(\frac{u}{a})+C,\qquad (a=1)$
$I=[\sinh^{-1}u]_{0}^{1}$
$=\sinh^{-1}1-\sinh^{-1}0$
$=\sinh^{-1}1-0$
$=\sinh^{-1}1$
$(b)$
Using the formulas in the box above these exercises, by which
$\sinh^{-1}x=\ln(x+\sqrt{x^{2}+1}),$
We would have
$\displaystyle \int\frac{du}{\sqrt{a^{2}+u^{2}}}=\sinh^{-1}(\frac{u}{a})+C=\ln[\frac{u}{a}+\sqrt{(\frac{u}{a})^{2}+1}]+C$
So, with $a=1,$
$I=[\ln(u+\sqrt{(u)^{2}+1}]_{0}^{1}$
$=\ln(1+\sqrt{1^{2}+1})-\ln(0+\sqrt{0^{2}+1})$
$=\ln(1+\sqrt{2})-\ln(1)$
$=\ln(1+\sqrt{2})$
