Question

how do i solve this? thanks!!

Answer 1

\[\int\limits_{0}^{1} \frac{ 1 }{ 1+\sqrt[3]{x} }dx\]

Answer 2

what do you know so far?

Answer 3

would u = 1 + 3sqrtx and du = dx?

Answer 4

This requires a u sub, what is the obvious u sub here?

Answer 5

you seem to know enough to tackle the problem

Answer 6

Try it out and see what you get!

Answer 7

You can also try \[u=\sqrt[3]{x}\] but then you probably will end up doing two substitutions, mess around with it and see what you get

Answer 8

ohh okay, err so would this be on the right track?
u = 1 + 3sqrtx and du = 3udu du = 3sqrtx

Answer 9

I'm still quite confused :( @Astrophysics :(

Answer 10

oops *dx=3sqrtx ? :/

Answer 11

\[\int \frac{dx}{1+\sqrt[3]{x}}\]\[u=x~,~ du=dx\]\[\int\frac{du}{1+u^{1/3}}\]

Answer 12

I left out the numbers

Answer 13

okay!! and so the next step would be the solve this integral? :/

Answer 14

you really want to sub \(u=x\) ?

Answer 15

wait, so we don't sub u=x? :/

Answer 16

what good does that do

Answer 17

im not sure :/ what would i make the sub be then? :/

Answer 18

unless u hate the letter `x` ..

Answer 19

\(\large\rm u=1+\sqrt[3]{x}\) or \(\large\rm u=\sqrt[3]{x}\)

I think both of these subs will work out just fine :)
Err no, go with the first one. I dunno bout that other guy, he gave me a funny look.

Answer 20

Where'd you leave off food man? :O
Where you stuck at?

Answer 21

lol im confused:/

so u = x ?

Answer 22

Oh what about changing this to arctan function..

Answer 23

\[\large\rm u=1+\sqrt[3]{x},\qquad\qquad du=\frac{1}{3x^{2/3}}dx\]Do you understand how to differentiate that u? :o

Answer 24

i think so :) what is next?

Answer 25

Now you have to apply a bunch of sneaky little tricks!

Answer 26

how do i do that?? :O

Answer 27

just differentiate what you have and try to finish it

Answer 28

Isolate the dx, that's one of the things we're substituting stuff in for after all.\[\large\rm 3x^{2/3}du=dx\]I'm gonna use rules of exponents to write it like this:\[\large\rm 3(x^{1/3})^2du=dx\]

Answer 29

that means \(\large x^\frac{1}{3} \)

Answer 30

We can sub something in for that x^(1/3) by using our equation involving u.

Answer 31

\(\large\rm u=1+x^{1/3}\quad\to\quad x^{1/3}=u-1\)

Answer 32

So there is our dx,\[\large\rm 3(u-1)^2du=dx\]

Answer 33

Was that super confusing? :o

Answer 34

Oh why did I miss this lol D'oh!

Answer 35

As I have said earlier, you have sufficient knowledge to tackle this problem. Just see to it that you finish what you started so you can learn different techniques along the way.

Answer 36

okay, haha yes a bit confusing but i think i follow.. sort of…. :P

what do i do next?

Answer 37

Substitute in your pieces,\[\large\rm \color{orangered}{1+\sqrt[3]{x}=u},\qquad\qquad \color{royalblue}{dx=3(u-1)^2du}\]\[\large\rm \int\limits \frac{1}{\color{orangered}{1+\sqrt[3]{x}}}\color{royalblue}{dx}=?\]

Answer 38

And from there, it shouldn't be too bad :)
You have to expand out the square in the numerator,
and you can divide each term by u, and integrate term by term.

Answer 39

best coaching ever

Answer 40

okay, er so do i get this? \[\frac{ 3x ^{2/3} }{ 2 } - 3\sqrt[3]{x} + 3\log(\sqrt[3]{x}+1) + C \]

Answer 41

@zepdrix ?

Answer 42

sec, doing some calculations :)

Answer 43

okie :)

Answer 44

What did you get after integrating in u?
I feel like you're missing a 2 in the middle maybe.\[\large\rm =3\left(\frac{1}{2}u^2-2u+\ln u\right)\]Something similar or no? :o
We can substitute back in a sec, I'm just curious if yours looks the same up to this point.

Answer 45

yes, i have that :

Answer 46

:)

Answer 47

When you get to this point, you normally have two options:
~find new boundary values for your integral, in terms of u, instead of x.
~or undo your substitution and use the original values for integrating.

I have a feeling... that with this monster it's going to be easier to get new boundaries for u as opposed to subbing back in those weird x's.\[\large\rm x=0 \qquad\to\qquad u=?\]\[\large\rm x=1\qquad\to\qquad u=?\]

Answer 48

ohh okay :) u=1 , u=2 ?

Answer 49

\[\large\rm =3\left(\frac{1}{2}u^2-2u+\ln u\right)_1^2\]Mmm sounds good!

Answer 50

so now i plug in? and i get

3((2-4+ln(2) -(1/2-2+ln(1)) ?

Answer 51

so 3(-2+ln(2) - 1/2 +2 -ln(1) = 3(-1/2 +ln(1) ?

Answer 52

Your ln(2) disappeared, uh oh

Answer 53

oh oops so it should be 3(-1/2 + ln(2) + ln(1)) ?

Answer 54

You could go a tad further with it if you wanted, ln(1)=0.
\[\large\rm =ln(2^3)-\frac{3}{2}\]
but whatever,
yay good job \c:/

Answer 55

ooh okay, so would that be my solution?

Answer 56

Yes, that or a decimal value, depending on which your teacher prefers.

Answer 57

ooh yay!! thanks so much!! to all of you!!:)