megannicole51
  • megannicole51
Find the integral of ln^5(z)/z
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
make a u substitution \[u=\ln(z), du =\frac{dz}{z}\] giving you \[\int u^5du\]
anonymous
  • anonymous
now i know you remember the power rule backwards ! \[\int u^5du=\frac{u^6}{6}\] right?!
megannicole51
  • megannicole51
yes i remember:D

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anonymous
  • anonymous
then replace \(u\) by \(\ln(z)\) and get \[\frac{1}{6}\ln^6(z)\]
anonymous
  • anonymous
and dont forget the stupid \(+C\) out at the end...
megannicole51
  • megannicole51
thank you :D i just need to keep practicing these
anonymous
  • anonymous
btw turning \(\int u^5du=\frac{u^6}{6}\) is the power rule backwards that u-substitution is the chain rule backwards
anonymous
  • anonymous
\[(f(g(x))'=f'(g(x))g'(x)\] \[\int f'(g(x))g'(x)dx=f(g(x))\]
anonymous
  • anonymous
of course you had to recognize \(\frac{1}{z}\) as the derivative of \(\ln(z)\)
megannicole51
  • megannicole51
i cant put ln into my online stuff so how would i be able to write ln^6 a different way?
megannicole51
  • megannicole51
(1/x)?
anonymous
  • anonymous
i am not sure exactly what you are asking
anonymous
  • anonymous
you have to put this in a computer homework? try (ln(z)^6)/6
megannicole51
  • megannicole51
my homework wont let me put in the ln^6 bc it doesnt recognize ln
anonymous
  • anonymous
you can't write "ln" you have to write "ln(z)"
anonymous
  • anonymous
like i said, try "(ln(z)^6)/6+C" that should work
anonymous
  • anonymous
just like you can't write "sin" you have to write "sin(x)"
megannicole51
  • megannicole51
perfect:) thank you!
megannicole51
  • megannicole51
would i be able to solve a problem if there is a square root in the denominator?
anonymous
  • anonymous
i bet it if is in the book or on line you can solve it you got a specific example?
megannicole51
  • megannicole51
its.....4e^4squareroot of y divided by square root y
anonymous
  • anonymous
\[\int\frac{ 4e^{4\sqrt{x}}}{\sqrt{x}}dx\]?
anonymous
  • anonymous
lets take that first 4 out front first and write \[4\int \frac{e^{4\sqrt{x}}}{\sqrt{x}}dx\]
megannicole51
  • megannicole51
okay:) easy enough lol
anonymous
  • anonymous
another u-substitution \(e^{4\sqrt{x}}\) is a composite function the "outside function " being \(e^u\) and the "inside function" being \(4\sqrt{x}\)
anonymous
  • anonymous
and it just so happens that the derivative of the inside function \(4\sqrt{x}\) is almost staring us right in the face the derivative of \(4\sqrt{x}\) is \(\frac{2}{\sqrt{x}}\)
anonymous
  • anonymous
you don't have \(\frac{2dx}{\sqrt{x}}\) you have \(\frac{dx}{\sqrt{x}}\) no matter, it is just a constant, we can adjust
anonymous
  • anonymous
ready ?
anonymous
  • anonymous
or did i lose you?
megannicole51
  • megannicole51
how did u get 2/root x from 4*root x?
megannicole51
  • megannicole51
jk i know what u did
anonymous
  • anonymous
i took the derivative of \(4\sqrt{x}\) and got \(\frac{2}{\sqrt{x}}\) ok? we need that, it is important
megannicole51
  • megannicole51
okay im good with that now:) proceed...
anonymous
  • anonymous
ok here goes put \(u=4\sqrt{x}, du =\frac{2}{\sqrt{x}}dx\) and so \(\frac{1}{2}du=\frac{dx}{\sqrt{x}}\)
anonymous
  • anonymous
making the substitution gives \[2\int e^udu\]
anonymous
  • anonymous
the 2 because we divided the 4 out front by the 2 in \(\frac{1}{2}du\)
anonymous
  • anonymous
now this is a rather easy integral right? since \(e^u\) is its own anti derivative and also its own anti derivative so \[2\int e^udu=2e^u\]
megannicole51
  • megannicole51
okay so far so good
anonymous
  • anonymous
ok we are done when we go back and replace \(u\) by \(4\sqrt{x}\) and get a final answer of \[2e^{4\sqrt{x}}+C\]
megannicole51
  • megannicole51
wait how did we get rid of the radical in the denominator?
anonymous
  • anonymous
your best hope of undersanding this gimmick is to take the derivative of \[2e^{4\sqrt{x}}\] and see that in fact you get \[\frac{4e^{4\sqrt{x}}}{\sqrt{x}}\]
anonymous
  • anonymous
we got rid of the radical in the denominator when we wrote \[\frac{1}{2}du=\frac{dx}{\sqrt{x}}\]
megannicole51
  • megannicole51
the only part i don't understand is that
megannicole51
  • megannicole51
where did the 1/2 come from
megannicole51
  • megannicole51
actually i see it now
anonymous
  • anonymous
here is what i would like you to do really, if you want to understand this trick take the derivative and see that you get what you want then you will understand the u substitution i can walk you through it if you like
megannicole51
  • megannicole51
sure:)
anonymous
  • anonymous
ok lets ignore the annoying 4 out front and just take the derivative of \[e^{4\sqrt{x}}\]
megannicole51
  • megannicole51
oh and just a quick question....so in my notes theres a problem e^(3x)/7+e^(3x) and it says i cant do it but im not sure why
anonymous
  • anonymous
by the chain rule, you get \[e^{4\sqrt{x}}\frac{d}{dx}[4\sqrt{x}]\] which is \[e^{4\sqrt{x}}\times \frac{2}{\sqrt{x}}\]
megannicole51
  • megannicole51
okay:) got it....keep join
megannicole51
  • megannicole51
*goin
anonymous
  • anonymous
well what is another way to write this? it is the same as \[\frac{2e^{4\sqrt{x}}}{\sqrt{x}}\]
megannicole51
  • megannicole51
okay i see it
anonymous
  • anonymous
so when you take the derivative of the composite function out pops the derivative of the inside function you are now trying to go backwards
anonymous
  • anonymous
so instead of it popping out, it pops in !
anonymous
  • anonymous
of course this trick only works if you are looking at \[f'(g(x))g'(x)\]
anonymous
  • anonymous
in other words, a composite function with the derivative of the inside function staring you in the face somewhere
megannicole51
  • megannicole51
that actually makes sense
anonymous
  • anonymous
imagine!
anonymous
  • anonymous
all this garbage is just tricks and gimmicks to find a function whose derivative is given largely a waste of time, but you have to get through it most of what you need is written as formulas on the back page of your text
megannicole51
  • megannicole51
yeah and i wish professors would teach you the easy tricks instead of making us suffer
anonymous
  • anonymous
math = suffer ok not really math + math teacher = suffer math = fun
anonymous
  • anonymous
hey, at least it is on line and you can cheat right?
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=int+4e^%284sqrt%28x%29%29%2Fsqrt%28x%29
megannicole51
  • megannicole51
omg i completely forgot about wolfram!
anonymous
  • anonymous
lol
anonymous
  • anonymous
now we will never see you here again will we?
anonymous
  • anonymous
could have saved your self like an hour at least but you did learn something i hope
megannicole51
  • megannicole51
nah i will have a lot more questions bc i like people showing me step by step....if i just had someone show me an example and just write out the steps usually i could figure it out
anonymous
  • anonymous
k see you later then good luck!
megannicole51
  • megannicole51
thank you so much for all of your help!

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