Question

indefinite integral u sub

Answer 1

\[\int\limits_{?}^{?}\frac{ (lnx)^{2}+2\ln(x^{2}-1) }{ x }\]

Answer 2

so what I did was break it up first

Answer 3

\[\int\limits_{?}^{?}\frac{ (lnx)^{2} }{ x }dx+\int\limits_{?}^{?}\frac{ 2\ln(x^{2}-1) }{ x }dx\]

Answer 4

and I can solve the first half but not the second

Answer 5

\[\int\limits_{}^{}\frac{2\ln(x-1)}{x} dx+\int\limits_{}^{}\frac{2 \ln(x+1)}{x} dx\]
don't know if this helps on the second part yet
but this something we can also do

Answer 6

x^2-1=(x-1)(x+1)
ln(ab)=ln(a)+ln(b)
is what i did

Answer 7

I don't think that is going to help us :(

Answer 8

me too

Answer 9

http://www.wolframalpha.com/input/?i=int%282+log_e%28x%5E2-1%29%2Fx+%2Cx%29
wolfram says it isn't elementary

Answer 10

So I wonder if this is the exact integral you started with

Answer 11

the sheet is unclear so I wasnt sure if it should be ln(x^2-1) or ln(x^2)-1. I went with the former but I can post a pic of the question

Answer 12

so I guess they didn't use ( )
I hate when they don't do that
it does make it confusing
I think we are going to have to go with the latter interpretation

Answer 13

\[\int\limits \frac{( \ln(x))^2+2 \ln(x^2)-1}{x} dx\]

Answer 14

yep that would work :)

Answer 15

just a simple
u = ln x
substitution

Answer 16

\[\int\limits \frac{ (\ln(x))^2}{x} dx+4 \int\limits_{}^{}\frac{\ln(x)}{x} dx-\int\limits_{}^{}\frac{1}{x} dx\]
yeah use @kidrah69 's suggest sub for both the first two

Answer 17

no need to even separate it....right ?

Answer 18

I think they would put parenthesis around that whole thing if they meant that whole thing but honestly I think for clarity purposes they should go ahead in throw in ( ) just in case

Answer 19

no need

Answer 20

\(\Large \int (u^2+4u-1) du\)

see if you get this :)
which is easy to integrate

Answer 21

I did because I thought it looked glamorous

Answer 22

don't forget to resubstitute back u = ln x

Answer 23

what happen to integration 1/x w.r.t x?

Answer 24

no sub needed for the last integral

Answer 25

isnt \[\int\limits \frac{ 1 }{ x }=lnx\]

Answer 26

yes

Answer 27

u missed dx and +c
but yes

Answer 28

well actually ln|x|+c

Answer 29

putting the bars around the x is "more correct "

Answer 30

correct \(\checkmark \)

more correct \(\huge \checkmark \)
:P

Answer 31

ok so I got\[\frac{ 1 }{ 3 }(lnx)^{3}+2(lnx)^{2}-lnx\]

Answer 32

+c
and thats absolutely correct! :)

Answer 33

Thanks. Would you be willing to help me with one more question?

Answer 34

wait
should u get that 2 ?

Answer 35

ok, yes

Answer 36

\[\int\limits\limits \frac{ (\ln(x))^2}{x} dx+4 \int\limits\limits_{}^{}\frac{\ln(x)}{x} dx-\int\limits\limits_{}^{}\frac{1}{x} dx \\ \int\limits_{}^{} (\ln(x))^2 d(\ln(x)) +4 \int\limits_{}^{} \ln(x) d(\ln(x)) -\ln|x|+C \\ \frac{(\ln(x))^3}{3}+4 \frac{(\ln(x))^2}{2}-\ln|x|+C\]

Answer 37

looks correct to me

Answer 38

just confirming :)

Answer 39

For this I expanded it using properties of logs and then took the derivative and got\[y'=\frac{ 4 }{ 2x+1 }+\frac{ 3 }{ x-1 }-\frac{ 6 }{ 2x-1 }-\frac{ 2 }{ x+1 }\] I wanted to know if there might be an easier way to combine this or if I just have to work through the algebra

Answer 40

you wanted to integrate y w.r.t x ?

Answer 41

\(\ln ab = \ln a + \ln b \)
\(\ln a^b = b \ln a \)

Answer 42

ok

Answer 43

if you were meant to differentiate y you did it correctly
do you really have to combine the fractions lol

Answer 44

I dont know. I dont think so since the next question is the same just without the ln so we take ln of both sides and do the same thing, but Im just not sure

Answer 45

so I wanted to ask

Answer 46

well no need to take ln of both sides
since there is ln just on that one side
but if you really want to combine the fractions I would start combining the fractions that have conjugate bottoms
\[y'=\frac{4}{2x+1}-\frac{6}{2x-1}+\frac{3}{x-1}-\frac{2}{x+1} \\ y'=\frac{4(2x-1)-6(2x+1)}{(2x+1)(2x-1)}+\frac{3(x+1)-2(x-1)}{(x-1)(x+1)}\]

Answer 47

painful

Answer 48

\[y'=\frac{8x-4-12x-6}{4x^2-1}+\frac{3x+3-2x+2}{x^2-1}\]

Answer 49

then simplify those tops
then combine those fractions

Answer 50

alright. Thank you