Jtdylktuy

So I have the following problem:

$$int_1^{e} frac{1}{xsqrt{1+ln^2x}}dx $$

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?

Hint. One may perform the change of variable
$$
u=ln x,qquad du=frac{dx}x,
$$ giving
$$
int frac{1}{xsqrt{1+ln^2x}}:dx=int frac{du}{sqrt{1+u^2}}
$$ then one may notice that
$$
left[ln left(u+ sqrt{1+u^2}right)right]'=frac{1+frac{u}{sqrt{1+u^2}}}{u+sqrt{1+u^2}}=frac{1}{sqrt{1+u^2}}.
$$

once we preform the change of variables $u=log x$, we of course have
$$I=intfrac{mathrm{d}u}{sqrt{1+u^2}}$$
Which can be computed using the following identity with hyperbolic trig. functions:
$$cosh^2t-sinh^2t=1$$
Substitution, baby: $u=sinh tRightarrow mathrm{d}u=cosh t mathrm{d}t$
$$I=intfrac{cosh t mathrm{d}t}{sqrt{1+sinh^2t}}$$
$$I=intfrac{cosh t mathrm{d}t}{sqrt{cosh^2t}}$$
$$I=intfrac{cosh t mathrm{d}t}{cosh t}$$
$$I=intmathrm{d}t$$
$$I=t$$
$$I=text{arcsinh},u$$
$$I=text{arcsinh},log x$$
Noting that $$text{arcsinh},x=logbig(sqrt{x^2+1}+xbig)$$
Of course provides the integral:
$$I=logbigg(sqrt{log^2x+1}+log xbigg)$$
QED

Remember that $log x$ is the natural logarithm.

Edit:

adding the absolute value bars in like so:
$$I=logbigg|sqrt{log^2x+1}+log xbigg|$$
Extends the domain of the anti-derivative, which is useful if required.

Edit 2.0:

plugging in the endpoints:
$$Ibig|_1^e =logbigg(sqrt{log^2e+1}+log ebigg)-logbigg(sqrt{log^21+1}+log 1bigg)$$
$$Ibig|_1^e =logbigg(sqrt{1+1}+1bigg)-logbigg(sqrt{0+1}+0bigg)$$
$$Ibig|_1^e =logbig(sqrt{2}+1big)$$
QED