Finding the definite integral $\int_1^e \frac{dx}{x\sqrt{1+\ln^2x}}$

Question

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?

Answer 1

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 $$ \left[\ln \left(u+ \sqrt{1+u^2}\right)\right]'=\frac{1+\frac{u}{\sqrt{1+u^2}}}{u+\sqrt{1+u^2}}=\frac{1}{\sqrt{1+u^2}}. $$

Answer 2

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