evaluate the integral lnx/(x*sqrt((lnx)^2+1) very confused

Question

ganeshie8 · Answer

You cannot break a radical... So whenever u see radical,
the first thing to try is to substitute the stuff inside radical

anonymous · Answer

yeah i set u = lnx^2+1

ganeshie8 · Answer

careful lnx^2 is very much different from (lnx)^2

anonymous · Answer

hmm this is what i got so far 1/2 integral 1/x * u^-1/2

anonymous · Answer

du/2=lnx

ganeshie8 · Answer

hold up

ganeshie8 · Answer

$\large  u =   (\ln x)^2 + 1 $

$\large  du =   2 (\ln x) (\ln x)'  dx$

$\large  du =   2 (\ln x) (\dfrac{1}{x}) dx $

$\large  \dfrac{du}{2} =   \dfrac{\ln x}{x} dx $

anonymous · Answer

oh chain rule right?

ganeshie8 · Answer

yes :)

anonymous · Answer

those always screw me over

ganeshie8 · Answer

plug them in the integrand
now it simplifies nicely

ganeshie8 · Answer

lol they does every one in the start...

ganeshie8 · Answer

$$\large   \int  \dfrac{   \ln x}{x*\sqrt{(\ln x)^2+1}} dx$$

after substitution becomes :

$$\large   \int  \dfrac{   1}{\sqrt{u}} \dfrac{du}{2}$$

ganeshie8 · Answer

which is trivial to integrate

anonymous · Answer

1/2 integral u^(-1/2) turns into 1/2 * 2u^(1/2) which then is u^1/2 then i plug u back in so it's sqrt((lnx)^2+1)?

ganeshie8 · Answer

looks good, and dont forget the arbitrary constant