Finding the definite integral $\int_1^e \frac{dx}{x\sqrt{1+\ln^2x}}$ So I have the following problem: 
$$\int_1^{e} \frac{1}{x\sqrt{1+\ln^2x}}dx $$
Can somebody comfirm that the integral of this is
$$\ln|\sqrt{1+\ln^2x}+ \ln x|+C$$
and I that the anwser is $$\ln |\sqrt{2}+1|$$
that is aproximately 0.8814
Does anyone else got the same anwser?
 A: Hint. One may perform the change of variable
$$
u=\ln x,\qquad du=\frac{dx}x,
$$ giving
$$
\int \frac{1}{x\sqrt{1+\ln^2x}}\:dx=\int \frac{du}{\sqrt{1+u^2}}
$$ then one may notice that
$$
\left[\ln \left(u+ \sqrt{1+u^2}\right)\right]'=\frac{1+\frac{u}{\sqrt{1+u^2}}}{u+\sqrt{1+u^2}}=\frac{1}{\sqrt{1+u^2}}.
$$
A: once we preform the change of variables $u=\log x$, we of course have
$$I=\int\frac{\mathrm{d}u}{\sqrt{1+u^2}}$$
Which can be computed using the following identity with hyperbolic trig. functions:
$$\cosh^2t-\sinh^2t=1$$
Substitution, baby: $u=\sinh t\Rightarrow \mathrm{d}u=\cosh t\ \mathrm{d}t$
$$I=\int\frac{\cosh t\ \mathrm{d}t}{\sqrt{1+\sinh^2t}}$$
$$I=\int\frac{\cosh t\ \mathrm{d}t}{\sqrt{\cosh^2t}}$$
$$I=\int\frac{\cosh t\ \mathrm{d}t}{\cosh t}$$
$$I=\int\mathrm{d}t$$
$$I=t$$
$$I=\text{arcsinh}\,u$$
$$I=\text{arcsinh}\,\log x$$
Noting that $$\text{arcsinh}\,x=\log\big(\sqrt{x^2+1}+x\big)$$
Of course provides the integral:
$$I=\log\bigg(\sqrt{\log^2x+1}+\log x\bigg)$$
QED
Remember that $\log x$ is the natural logarithm.
Edit:
adding the absolute value bars in like so:
$$I=\log\bigg|\sqrt{\log^2x+1}+\log x\bigg|$$
Extends the domain of the anti-derivative, which is useful if required.
Edit 2.0:
plugging in the endpoints:
$$I\big|_1^e =\log\bigg(\sqrt{\log^2e+1}+\log e\bigg)-\log\bigg(\sqrt{\log^21+1}+\log 1\bigg)$$
$$I\big|_1^e =\log\bigg(\sqrt{1+1}+1\bigg)-\log\bigg(\sqrt{0+1}+0\bigg)$$
$$I\big|_1^e =\log\big(\sqrt{2}+1\big)$$
QED
