Answer 1

Drawings wont work in the question :)

Answer 2

eish k let me write it then

Answer 3

ok

Answer 4

Determine integral (with top part of it with e^3 and the bottom part of it e^2) of: In(In w)/wlog3w (this 3 its small and just slight below log) is

Answer 5

You can draw or use the Equation editor While replying :)
Hope someone will help you out!

Answer 6

\[\int\limits_{e^2}^{e^3}In(Inw)/wlog3w dw\]

Answer 7

i tried to type it

Answer 8

but the 3 in log is slight below the log, its not 3w, the 3 its slightly belowthe log

Answer 9

oh ok so log base 3

Answer 10

yes

Answer 11

ok first get rid of that log by using change of base
\[\rightarrow \log_{3}w = \frac{\ln w}{\ln 3} \]
simplify the integral
\[\rightarrow \ln 3\int\limits_{?}^{?}\frac{\ln (\ln w)}{w \ln w}dw\]

Answer 12

then make substitution
u = ln w
du = 1/w dw --> dw = w du

\[\rightarrow \ln 3\int\limits_{?}^{?} \frac{\ln u}{u}du\]

Answer 13

then use integration by parts:

x = ln u dv = 1/u
dx = 1/u v = ln u

\[\int\limits_{?}^{?} \frac{\ln u}{u}du = \ln^{2}u - \int\limits_{?}^{?} \frac{\ln u}{u}du\]
\[2\int\limits_{?}^{?} \frac{\ln u}{u}du = \ln^{2} u\]
\[\int\limits_{?}^{?} \frac{\ln u}{u}du = \frac{\ln^{2} u}{2}\]
sub back in for u and put limits in
\[\rightarrow \ln 3(\frac{\ln^{2} (\ln w)}{2}) from e^{3} \to e^{2}\]

Answer 14

i get about 0.4

Answer 15

thank you i see, i never knew about intergration by parts

Answer 16

do you mind explaining to me how it works

Answer 17

welcome, sure

anytime you have a product of 2 functions, you can use integration by parts
\[\int\limits_{}^{}f(x)g(x) dx\]
assign either f or g as u and the other as dv, where u and v are functions of x
this leads to the equation:
\[\int\limits_{}^{}u dv = uv -\int\limits_{}^{}v du\]
so to use the equation you must find du and v
lets say u = f(x) then du = f'(x)
then dv = g(x) and v = integral g(x)

Answer 18

here is an example:
\[\int\limits_{}^{}xe^{x}dx\]
u = x dv = e^x
du = dx v = e^x
\[\int\limits\limits_{}^{}xe^{x}dx = xe^{x}-\int\limits_{}^{}e^{x} dx\]

Answer 19

k even if both function are are in the division format? e.g. \[e^xdivx\]

Answer 20

u can still use the the formula?

Answer 21

yes, but one of the functions would be 1/x

u = e^x dv = 1/x
du = e^x v = ln x

notice this will not help much though
actually integral e^x/x can't be done using regular methods but in general yes even division can be represented as product of 2 functions

Answer 22

oh k see i get it, thank you, Be blessed