Find the integral of ln^5(z)/z

- megannicole51

Find the integral of ln^5(z)/z

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- schrodinger

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- anonymous

make a u substitution
\[u=\ln(z), du =\frac{dz}{z}\] giving you
\[\int u^5du\]

- anonymous

now i know you remember the power rule backwards !
\[\int u^5du=\frac{u^6}{6}\] right?!

- megannicole51

yes i remember:D

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- anonymous

then replace \(u\) by \(\ln(z)\) and get
\[\frac{1}{6}\ln^6(z)\]

- anonymous

and dont forget the stupid \(+C\) out at the end...

- megannicole51

thank you :D i just need to keep practicing these

- anonymous

btw turning \(\int u^5du=\frac{u^6}{6}\) is the power rule backwards
that u-substitution is the chain rule backwards

- anonymous

\[(f(g(x))'=f'(g(x))g'(x)\]
\[\int f'(g(x))g'(x)dx=f(g(x))\]

- anonymous

of course you had to recognize \(\frac{1}{z}\) as the derivative of \(\ln(z)\)

- megannicole51

i cant put ln into my online stuff so how would i be able to write ln^6 a different way?

- megannicole51

(1/x)?

- anonymous

i am not sure exactly what you are asking

- anonymous

you have to put this in a computer homework?
try (ln(z)^6)/6

- megannicole51

my homework wont let me put in the ln^6 bc it doesnt recognize ln

- anonymous

you can't write "ln"
you have to write "ln(z)"

- anonymous

like i said, try "(ln(z)^6)/6+C" that should work

- anonymous

just like you can't write "sin" you have to write "sin(x)"

- megannicole51

perfect:) thank you!

- megannicole51

would i be able to solve a problem if there is a square root in the denominator?

- anonymous

i bet it if is in the book or on line you can solve it
you got a specific example?

- megannicole51

its.....4e^4squareroot of y divided by square root y

- anonymous

\[\int\frac{ 4e^{4\sqrt{x}}}{\sqrt{x}}dx\]?

- anonymous

lets take that first 4 out front first and write
\[4\int \frac{e^{4\sqrt{x}}}{\sqrt{x}}dx\]

- megannicole51

okay:) easy enough lol

- 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

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

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

ready ?

- anonymous

or did i lose you?

- megannicole51

how did u get 2/root x from 4*root x?

- megannicole51

jk i know what u did

- anonymous

i took the derivative of \(4\sqrt{x}\) and got \(\frac{2}{\sqrt{x}}\) ok? we need that, it is important

- megannicole51

okay im good with that now:) proceed...

- 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

making the substitution gives
\[2\int e^udu\]

- anonymous

the 2 because we divided the 4 out front by the 2 in \(\frac{1}{2}du\)

- 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

okay so far so good

- 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

wait how did we get rid of the radical in the denominator?

- 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

we got rid of the radical in the denominator when we wrote
\[\frac{1}{2}du=\frac{dx}{\sqrt{x}}\]

- megannicole51

the only part i don't understand is that

- megannicole51

where did the 1/2 come from

- megannicole51

actually i see it now

- 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

sure:)

- anonymous

ok lets ignore the annoying 4 out front and just take the derivative of
\[e^{4\sqrt{x}}\]

- 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

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

okay:) got it....keep join

- megannicole51

*goin

- anonymous

well what is another way to write this? it is the same as
\[\frac{2e^{4\sqrt{x}}}{\sqrt{x}}\]

- megannicole51

okay i see it

- 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

so instead of it popping out, it pops in !

- anonymous

of course this trick only works if you are looking at
\[f'(g(x))g'(x)\]

- anonymous

in other words, a composite function with the derivative of the inside function staring you in the face somewhere

- megannicole51

that actually makes sense

- anonymous

imagine!

- 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

yeah and i wish professors would teach you the easy tricks instead of making us suffer

- anonymous

math = suffer
ok not really
math + math teacher = suffer
math = fun

- anonymous

hey, at least it is on line and you can cheat right?

- anonymous

http://www.wolframalpha.com/input/?i=int+4e^%284sqrt%28x%29%29%2Fsqrt%28x%29

- megannicole51

omg i completely forgot about wolfram!

- anonymous

lol

- anonymous

now we will never see you here again will we?

- anonymous

could have saved your self like an hour at least
but you did learn something i hope

- 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

k see you later then
good luck!

- megannicole51

thank you so much for all of your help!

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