mjhjmtu

So I have the following problem:

$$int_1^e frac1xsqrt1+ln^2xdx $$

Can somebody comfirm that the integral of this is

$$ln|sqrt1+ln^2x+ ln x|+C$$

and I that the anwser is $$ln |sqrt2+1|$$
that is aproximately 0.8814

Does anyone else got the same anwser?

Hint. One may perform the change of variable
$$
u=ln x,qquad du=fracdxx,
$$ giving
$$
int frac1xsqrt1+ln^2x:dx=int fracdusqrt1+u^2
$$ then one may notice that
$$
left[ln left(u+ sqrt1+u^2right)right]'=frac1+fracusqrt1+u^2u+sqrt1+u^2=frac1sqrt1+u^2.
$$

once we preform the change of variables $u=log x$, we of course have
$$I=intfracmathrmdusqrt1+u^2$$
Which can be computed using the following identity with hyperbolic trig. functions:
$$cosh^2t-sinh^2t=1$$
Substitution, baby: $u=sinh tRightarrow mathrmdu=cosh t mathrmdt$
$$I=intfraccosh t mathrmdtsqrt1+sinh^2t$$
$$I=intfraccosh t mathrmdtsqrtcosh^2t$$
$$I=intfraccosh t mathrmdtcosh t$$
$$I=intmathrmdt$$
$$I=t$$
$$I=textarcsinh,u$$
$$I=textarcsinh,log x$$
Noting that $$textarcsinh,x=logbig(sqrtx^2+1+xbig)$$
Of course provides the integral:
$$I=logbigg(sqrtlog^2x+1+log xbigg)$$
QED

Remember that $log x$ is the natural logarithm.

Edit:

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