evaluate the integral integral of (ln(4x))^2dx
im assuming this is integration by parts. but i am not sure what to make my u and what to make dv
would ln(4x) be good for u? and i would use chain rule to find derivative?
\[\int\limits (\ln(4x))^2 dx\] is it like this?
yes it is
You can just let u = (ln(4x))^2 and dv = 1
hey astrophysics can you please help me I posted a question
ok! where do you get the 1 from?
pls help i need to get rest
We can think of there being a 1 when we have such functions dv=dx, it's like if you have to integrate tanx dx you would use by parts, lets u = tanx, and dv = dx
I guess it would've been better if I said dv = 1*dx
ohh yes i see! thank you
astro pls help
Not tanx I meant arctanx haha
hmm im not getting the right answer.. is it correct to say that the integral of ln4 is = to xln4 ?
\[(\ln(4x))^{2}x-\int\limits x \frac{ 2\ln4 }{ x } dx\]
so then i cancel out the x on the outside and the x in the denominator..
and i am left with the integral of 2ln4, so then i take out the 2 and put it behind the integral, and i am left with the integral of ln4
from there i took the integral of ln4 which i believe is xln4 and i put both parts together but the answer is incorrect
\[(\ln(4x))^2x-2xln4\] that is my answer..
but the answer says incorrect..
the answer is \[x(\ln (4x))^2-2xln (4x)+2x\] recheck once again :)
You're almost there.
thank you.. i am not seeing where the 2x at the end comes from.. ?
that is my work so far
u=4x \[\int\limits\frac{ (\ln (u)) }{ 4}du \]after some vry bad calculations nd simplifications u get- \[\frac{ 1 }{ 4 }((u)(\ln (u))^2 -2\ln (u)+2(u))\]and then when u simplify nd put u=4x then u get that answer..
ok i see, thank you!
no prblm but that is a long method..
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