Question

solve te question ill post to down (third one)

Answer 1

\[\int\limits_{}\sqrt[3]{x}*lnx^{2}*dx \]

Answer 2

This one looks like integration by parts. I don't know if you got there yet Korcan.

Answer 3

well i knomw the formula (just learning everthing by myself...) \[\int\limits_{} u* dv = uv - \int\limits_{} vdu\]

Answer 4

I see you have been doing a lot of work. Use ln x^2 as your u.

Answer 5

well ok let me...

Answer 6

well will it be like : \[ux \int\limits_{}\sqrt[3]{x}*x*du \]
if we take \[ u = \ln x^{2}\]

Answer 7

whops wiat there is something wrong with that. let me fix it

Answer 8

\[ux - \int\limits_{}\sqrt[3]{x}xdu\]

Answer 9

am i on true way?

Answer 10

You are doing good. I haven't done integration by parts in a long time. Try to follow this, remember ln x^2 is same as log x^2 http://www.wolframalpha.com/input/?i=integral+ [cube+root+%28x%29]ln+x^2

Answer 11

thnaks ^^

Answer 12

îll do i by my hand ^^ but now i have to go for dinner bb

Answer 13

What was for dinner.
u= ln x^2
du=2/x
v=x^(1/3)
dv=(1/[3x^(2/3)]

Answer 14

\[\int\limits_{}^{}u dv=uv -\int\limits_{}^{}v du\]

Answer 15

\[x ^{1/3}\ln x ^{2}-\int\limits_{}^{}2/x ^{2/3}\]Try to work from here. That is your answer.

Answer 16

mhmh can you explain me i dun got the process for dinner :D :D)

Answer 17

ahh okay now i got it by how can we say dv = (1/[3x^(2/3)] ???

Answer 18

Read up above, you have values of u, du, v, dv. Plug in formula. The right hand side of equation is the answer. You just need to integrate (2/x^2/3)

Answer 19

there are no more dx yea? ( i mean as a value ^^ ) ^^

Answer 20

wow without reading all of this the first thing i would do is write
\[\ln(x^2)=2\ln(x)\] and pull the 2 out of the whole thing. yes?

Answer 21

make my live easier i think

Answer 22

^^ but this way we have to result ^^

Answer 23

Would that help, if you go integration by parts?

Answer 24

let me give it a go
\[\int udv=uv-\int vdu\]
\[u=ln(x)\]
\[du=\frac{1}{x}dx\]
\[dv=\sqrt[3]x\]
\[v=\frac{2}{3}\sqrt[3]{x^4}\]

Answer 25

so far so good?

Answer 26

looks like nice ^^

Answer 27

but the way we did is looks like easier

Answer 28

I was wondering, can you re-insert the 2 (from ln x) at the end of this process?

Answer 29

if so we get
\[\frac{2}{3}\sqrt[3]{x^4}\ln(x)-\frac{2}{3}\int \sqrt[3]{x}dx\]

Answer 30

yeah 2 comes outside of whole thing. right out in front of the integral sign

Answer 31

i have been ignoring it

Answer 32

but i made a mistake!

Answer 33

\[dv=\sqrt[3]{x}dx\]
\[v=\frac{4}{3}\sqrt[3]{x^4}\]

Answer 34

sorry about that

Answer 35

no even that is not right!

Answer 36

lordamercy. hope you are not writing this down. i should have done it on paper first. sorry

Answer 37

once more
\[dv=\sqrt[3]{x}dx\]
\[v=\frac{3}{4}\sqrt{x^4}\] that one i believe is correct finally

Answer 38

whew. so we get
\[\frac{3}{4}\sqrt{x^4}\ln(x)-\frac{3}{4}\int \sqrt[3]{x}dx\]

Answer 39

\[=\frac{3}{4}\sqrt[3]{x^4}-\frac{9}{16}\sqrt[3]{x^4}\] and now multiply by 2 to get the "final answer"

Answer 40

Thank you, Good work.

Answer 41

yes GOOD WORK STELİTE Thanks much ^^

Answer 42

no not good work yet

Answer 43

yes that is it. once you multiply by 2 you have it!

Answer 44

I still don't understand how it is valid to pull through the 2. After all, it is not just an integral sign, it is an equation.

Answer 45

it is a fact that
\[\int cf(x)dx = c\int f(x)dx\]

Answer 46

since
\[\frac{d}{dx} cf(x)=c\frac{d}{dx}f(x)\]

Answer 47

it is also a fact that
\[\ln(x^2)=2\ln(x)\] but besides that think of the step where you integrate
\[\int\frac{3}{4}\sqrt[3]{x}dx\] you just write
\[\frac{3}{4}\int \sqrt[3]{x}dx\] without hesitation

Answer 48

or even
\[\int 5x^2dx=\frac{5}{3}x^3\] because the constant comes right through the integral sign.

Answer 49

Thanks for your. Korcan, you got it?

Answer 50

well i am trying to got it :)