Question

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

Answer 1

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

Answer 2

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

Answer 3

yes i remember:D

Answer 4

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

Answer 5

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

Answer 6

thank you :D i just need to keep practicing these

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

(1/x)?

Answer 12

i am not sure exactly what you are asking

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

perfect:) thank you!

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

okay:) easy enough lol

Answer 25

another u-substitution
\(e^{4\sqrt{x}}\) is a composite function
the "outside function " being \(e^u\) and the "inside function" being \(4\sqrt{x}\)

Answer 26

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 \(\frac{2}{\sqrt{x}}\)

Answer 27

you don't have \(\frac{2dx}{\sqrt{x}}\) you have \(\frac{dx}{\sqrt{x}}\) no matter, it is just a constant, we can adjust

Answer 28

ready ?

Answer 29

or did i lose you?

Answer 30

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

Answer 31

jk i know what u did

Answer 32

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

Answer 33

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

Answer 34

ok here goes
put \(u=4\sqrt{x}, du =\frac{2}{\sqrt{x}}dx\) and so \(\frac{1}{2}du=\frac{dx}{\sqrt{x}}\)

Answer 35

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

Answer 36

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

Answer 37

now this is a rather easy integral right? since \(e^u\) is its own anti derivative and also its own anti derivative
so
\[2\int e^udu=2e^u\]

Answer 38

okay so far so good

Answer 39

ok we are done when we go back and replace \(u\) by \(4\sqrt{x}\) and get a final answer of
\[2e^{4\sqrt{x}}+C\]

Answer 40

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

Answer 41

your best hope of undersanding this gimmick is to take the derivative of
\[2e^{4\sqrt{x}}\] and see that in fact you get
\[\frac{4e^{4\sqrt{x}}}{\sqrt{x}}\]

Answer 42

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

Answer 43

the only part i don't understand is that

Answer 44

where did the 1/2 come from

Answer 45

actually i see it now

Answer 46

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

Answer 47

sure:)

Answer 48

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

Answer 49

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

Answer 50

by the chain rule, you get
\[e^{4\sqrt{x}}\frac{d}{dx}[4\sqrt{x}]\] which is
\[e^{4\sqrt{x}}\times \frac{2}{\sqrt{x}}\]

Answer 51

okay:) got it....keep join

Answer 52

*goin

Answer 53

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

Answer 54

okay i see it

Answer 55

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

Answer 56

so instead of it popping out, it pops in !

Answer 57

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

Answer 58

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

Answer 59

that actually makes sense

Answer 60

imagine!

Answer 61

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

Answer 62

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

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

omg i completely forgot about wolfram!

Answer 67

lol

Answer 68

now we will never see you here again will we?

Answer 69

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

Answer 70

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

Answer 71

k see you later then
good luck!

Answer 72

thank you so much for all of your help!