Question

Take the Integral of...

Answer 1

\[\int\limits \frac{lnx}{x\sqrt{1+lnx^2}}dx\]

Answer 2

I got \[\ln (lnx + \sqrt{1+lnx^2} +C\]
is this correct?

Answer 3

Is it:
\[\sqrt{1+\ln(x^2)}\]
or
\[\sqrt{1+\ln^2(x)}\]

Answer 4

the bottom one

Answer 5

what was your procedure to obtain your answer?

Answer 6

from \( \displaystyle \int \frac{ \ln x }{x \sqrt{ 1 + \ln ^2 x }} \ dx \)

to \( \ln ( \ln x + \sqrt{ 1 + \ln^2 x } ) + C \)

Answer 7

Can you write it out, step by step?

Answer 8

let u = lnx
du =1/x dx
= \[\int\limits \frac{u}{\sqrt{1+u^2}}du\]

Answer 9

Looks like you guys have got it, but I just did it and I'd just like to say this integral works out beautifully.

Answer 10

that step looks good. what next? :)

Answer 11

It's yours access :)

Answer 12

then I just used that integral rule...
\[\int\limits \frac{xdx}{\sqrt{x^2+a^2}} = \ln(x+ \sqrt{x^2+a^2})\]

Answer 13

do I have to show the trig substitution?

Answer 14

i cannot rightfully say i have seen that identity. i was thinking of another "u" sub

Answer 15

That isn't quite the right identity. Just like access said, there is an alternative u substitution that you can use as well.

Answer 16

not quite like that... notice the numerator is x dx, the denominator inside the radical is x^2 + 1. the numerator is just a derivative of that radicand right?

Answer 17

acutally let u=tan(δ)

then du = sec^2δdδ

Answer 18

so now I get like

\[\int\limits \frac{\tanδ \sec^2δdδ}{|\secδ|}\]

Answer 19

i think you do get the correct answer if you simplify this and follow through. it just seems to take more work than v=u^2+1, dv=2u du

Answer 20

how do I get rid of the square root sign?

Answer 21

from 1/sqrt(v) dv is just power rule: 1/sqrt(v) = 1/v^(1/2) = v^(-1/2)

Answer 22

When I did this, I went like this:
\[\int\limits \frac{lnx}{x \sqrt{1+\ln^2(x)}}dx\]
\[u = 1 + \ln^2(x), du= 2\frac{lnx}{x}dx\]
\[\frac{1}{2}\int\limits \frac{1}{\sqrt{u}}du\]

Answer 23

oh I see I get it I got

\[1/2\int\limits dv/v\]

Answer 24

actually sqrtv

Answer 25

the v = u^2 + 1 part goes inside the square root
yeah.

Answer 26

=-1/21/1+u^2 = -1/2(1+ln^2(x)

Answer 27

i don't think you added to the power correctly. the square root will still be there
\( \int v^{-1/2} \implies v^{-1/2 \color{green}{+1}} / (-1/2+\color{green}{1}) \)

Answer 28

yea that's right haha

Answer 29

not to say that your initial approach was necessarily wrong, it just ends up with an answer like:
sec t t = arctan u
sec arctan u u = ln x
sec (arctan (ln x)) <-- now you'd ideally want to put this in terms of ln x using logic of compositions of trig/inverse trig and you'd eventually get sqrt(ln^2 x + 1)/1

Answer 30

oh I see but much more effort required...

Answer 31

You can also see in the integral tables in the back of most Calculus texts that:
\[\int\limits \frac{x}{\sqrt{a^2 + x^2}}dx=\sqrt{a^2+x^2} +C\]

Answer 32

ok yea that was what I think I was thinking of at the start.