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i want to confirm the answer of... integrate lnx/x dx

Mathematics
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my answer is ln lxl +c
let u=lnx du=dx/x \[\int\limits_{}^{}u du=\frac{u^2}{2}+C=\frac{(lnx)^2}{2}+C\] we can even check this: \[(\frac{(lnx)^2}{2}+C)'=2*\frac{lnx}{2}*\frac{1}{x}+0=\frac{lnx}{x}\]
do you got it?

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Other answers:

yes
k :)
q. how do i know what to choose?
i was looking for \[\int\limits_{}^{}f(x)*f'(x)dx\]
sometimes its not always that easy though
someone said it is like an art and i agree
jajaj..ok
it is not always easy to know what substititon to make but since it was in the form of above it was easy let u=f(x) du=f'(x) dx
\[\int\limits_{}^{}\frac{lnx}{x}dx=\int\limits_{}^{}lnx*\frac{1}{x} dx\]
so it was easy for me to notice pretty fast that (lnx)'=1/x so thats why i let u=lnx
ok..perfect
so how can i integrate tanx dx..... even though i know thats it is equal to -lnlcosxl+c
\[\int\limits_{}^{}\frac{sinx}{cosx} dx\] right?
tanx=sinx/cosx agree?
ohhhhhhhh... ok
yes
so that du/u
now what do you think the substitution will be?
with the negative sign of the sin
u=cos
du=sin x
sorry du=-sinx
du=-sinx dx (i like not to forget about this dx part)
oh yea
so show me what that gives us don't integrate yet
then u have integral of du/u
negative or positive?
- integral of du/u
yes and now integrate
so my result will be -ln/u/+c
-ln/cosx/+c
yes :)
yeaaa thank uuuu
np
so whats this \[\int\limits_{}^{}-tanx dx\] equal to?
ln/cosx/+c
becuase the negative sign will change to postive once u take the derivative of cosx
why u pic u =sinx?
accident
u=cosx du=-sinx dx
:)
good....:
you want to try one i made up? its not hard
ok
\[\int\limits_{}^{}\frac{x+1}{x^2+2x}dx\]
ok.. let me try
u=x^2 + 2x du=2x+2 dx =2(x+1) dx so, 1/2du=x+1 dx
very good so far
1/2 integral of du/u =1/2 lnlul+c =1/2 ln/x^2+2xl+c
:)
yeaaaaaaaaaaaaa
i love this stuff
lol it is nice
\[\int\limits_{}^{}\frac{f'(x)}{f(x)}dx\] so when we have this what will out answer be in the form of?
f(x)*f'(x)
look at the one you just did didn't we have f'/f dx?
ohh... sorry.. yes yes
=ln|f(x)|+C
yes
\[\int\limits_{}^{}[f(x)]^n*f'(x)dx\] \[n \neq -1\] what about this? what will our answer be?
i will give you a hint let u=f(x) so du=f'(x) dx
jjajjjaj... nice hint
where are you from? you say jajaja alot lol
dominican republic
i am just laughing
jajaaja= laughing?
yess... how can i laugh??
its fine the way you type your laughter no worries i like it
lol
so this all makes a little more sense?
the integral stuff?
why?
im asking do you get the integral stuff better?
ohhhhhh,, yess
good :) im fixing to go to sleep goodnight
for me is mornig.. but good nigth,,,
good morning* i didn't sleep all night lol
jajajajaaja
ayayay.. no good
thank u very much...
np if you want to try another this is a good one \[\int\limits_{}^{}x*\sqrt{x+1} dx\] the answer is given above in one of the other threads if you want to check yourself
later
ok,, thanks

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