Question

I need help with another problem with integrals

Answer 1

\[\int\limits_{}^{}\frac{ \ln(25) }{ (5-x)(\log_{5} (5-x))^3 }dx\]

Answer 2

Oh fun little u-sub it looks like :o

Answer 3

Ohhh it's log base 5, that's fancy.

Answer 4

I simplified log_5(5-x)^3 into ((ln(5-x)/(ln5))^3

Answer 5

then u = ln(5-x) , du = 1/5-x

Answer 6

Let's split some stuff up a sec,
\[\Large\rm =\ln(25)\int\limits \frac{1}{\left[\log_5(5-x)\right]^3}\left(\frac{1}{5-x}dx\right)\]We want \(\Large\rm u=\log_5(5-x)\).

Oh you did the base change BEFORE u-sub? mm I lik ethat better actually :)

Answer 7

So maybe factor the denominator of your base change out of the problem, might make it easier to look at:\[\Large\rm =\frac{\ln(25)}{\ln^3(5)}\int\limits\limits \frac{1}{\left[\ln(5-x)\right]^3}\left(\frac{1}{5-x}dx\right)\]

Answer 8

Looks like you're on the right track!

Answer 9

Woops, I didn't flip that correctly >.< My bad.

Answer 10

I thought its ln(25)(ln^3(5))

Answer 11

\[\Large\rm \ln(25)\ln^3(5)\int\limits\limits\limits \frac{1}{\left[\ln(5-x)\right]^3}\left(\frac{1}{5-x}dx\right)\]Yah good call c:

Answer 12

After you u-sub, looks like you can simply power rule, yes?

Answer 13

I get Integral of 1/u^3 yep

Answer 14

I got rid of the "cubed" and put it in front of the Ln

Answer 15

so it would be (ln25)(ln^3(5))[-1/(2(ln(5-x))^2)) +C

Answer 16

@alekos Nooo you can't do that :O\[\Large\rm \ln(5^3)\ne \ln^3(5)\]We were using fancy notation for the exponent.
It's not in the proper form to apply that log rule :(

Answer 17

Yeah its ln^3(5) = (ln5)^3

Answer 18

Mmm yah looks good!
If you want to put it back in terms of log base 5, it would require a couple tricky steps. Over simplifying usually isn't necessary though.

Answer 19

nvm I read something else

Answer 20

my final answer is Ln25/3Ln5 [Ln(Ln(5-x))]

Answer 21

:P

Answer 22

\[\frac{ -\ln(25)(\ln^3(5)) }{ 2\ln^2(5-x) }\]

Answer 23

No idea how I got that negative sign, can anyone fix me?

Answer 24

I got it after antideriving 1/u^3

Answer 25

Antideriving 1/u^3 = -1/2 u^-2

Answer 26

Oh oh oh I see it.

Answer 27

\[\Large\rm du=\ln(5-x) \qquad\to\qquad u\ne \frac{1}{5-x}dx\]

Answer 28

Don't forget to chain rule!

Answer 29

u=ln(5-x) , du= 1/5-x

Answer 30

I wrote my u and du backwards, sorry lolol

Answer 31

\[\Large\rm u=\ln(5-x) \qquad\to\qquad du\ne \frac{1}{5-x}dx\]

Answer 32

wolfram is a liar lol. I was right with my calculations then

Answer 33

\[\Large\rm du=\frac{1}{5-x}(5-x)'dx\]

Answer 34

du= -1/(5-x) dx

Answer 35

i see where i went wrong

Answer 36

any clue with wolfram doesn't do the chain rule? http://www.wolframalpha.com/input/?i=derivative+ln%285-x%29

Answer 37

ln(25) and ln^3(5) are constants so you simply take them out

Answer 38

wolfram is fine. just expresses -1/(5-x) as 1/(x-5)

Answer 39

\[\Large\rm -(5-x)=x-5\]

Answer 40

oh lol I didn't see it

Answer 41

well done Mateaus

Answer 42

thanks