Integral Calculus (Indefinite) what pointers should i think to choose the correct u.
in u-substitution

- anonymous

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- zepdrix

You're looking for a suitable `u` and `u'` to replace all of your integral.\[\Large \int\limits 3x^2\sqrt{x^3+4}\;dx\]Sometimes it helps to group things.
This is how I would group them.\[\Large \int\limits\sqrt{x^3+4}\left(3x^2\;dx\right)\]

- zepdrix

In the example:\[\Large u=x^3+4 \qquad\qquad\qquad du=\left(3x^2\;dx\right)\]

- zepdrix

\[\Large \int\limits\limits\sqrt{x^3+4}\left(3x^2\;dx\right) \qquad\to \qquad \int\limits \sqrt{u}(du)\]

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## More answers

- zepdrix

It's really takes a bit of practice to recognize what is going on.
Like you'll get a lot of weird looking stuff thrown at you.
Sometimes it's hard to identify your du.
Example:\[\Large \int\limits\frac{\ln x}{x}dx\]Another problem where I think grouping helps.\[\Large \int\limits\ln x\left(\frac{1}{x}dx\right)\]

- zepdrix

So in this example:\[\Large u=\ln x \qquad\qquad\qquad du=\left(\frac{1}{x}dx\right)\]

- zepdrix

It's not always that straight forward though.
Sometimes you need to apply a `u-sub` and then mess around with the pieces.
Example:\[\Large \int\limits \frac{x}{\sqrt{x-2}}dx\]

- zepdrix

In this example:\[\Large u=x-2 \qquad\qquad\qquad du=dx\]But we still need to deal with the x on top right?
So we'll solve our initial substitution for x.\[\Large u=x-2 \qquad\to\qquad x=u+2\]

- zepdrix

\[\Large \int\limits\limits \frac{x}{\sqrt{x-2}}dx \qquad=\qquad \int\limits \frac{u+2}{\sqrt u}du\]This is much easier to solve in u now.

- zepdrix

Do you have any specific examples you need help with? :o
Or any clarification on what I did above?

- dumbcow

@zepdrix , very thorough examples :)

- anonymous

cool now i get it its clear :), but what about the e^x and e^3x should i use the whole e^x as u? or just x?

- zepdrix

For a problem like this?\[\Large \int\limits e^{3x}dx\]Or something a bit more complex?

- anonymous

something like this \[\int\limits \frac{ e ^{2x} }{ (1+e ^{x}) ^{1/3}} dx\]

- anonymous

u see there are 2 \[e ^{x}\] but the other one is \[e ^{2x}\] how do i deal with this?

- zepdrix

mm sec XD thinking hehe

- zepdrix

Ok you have to think back to your rules of exponents.
We're allowed to rewrite the numerator like this,\[\Large e^{2x}\quad=\quad (e^x)^2\]

- zepdrix

I'm gonna do my fun grouping thing again and maybe it will jump out at you.
If not, that's ok. This one is a bit tricky.

- anonymous

ok im at it, ill do my best

- zepdrix

\[\Large \int\limits\frac{e^x}{ (1+e ^{x}) ^{1/3}}\left(e^x\;dx\right)\]

- zepdrix

So I split up the square which allowed me to group it a little weird.

- anonymous

haha never thought of doin that

- anonymous

but what sould i use as a U

- zepdrix

Let's try letting the inside of our denominator be our u and see what happens.
We might hit a bump in the road, if we do just have to think for a sec :)
\(\Large u=1+e^x\)

- zepdrix

So what do you get for du? :x

- anonymous

e^x

- zepdrix

\[\Large du=(e^x\;dx)\]Ok good.
Hmm looks like we still need to deal with that e^x in the numerator.
Any ideas? :)
Look back at the last example I gave, with the sqrt in the denominator!! :O

- anonymous

\[\int\limits \frac{(du)(du) }{(u)^{1/3} } dx\]

- anonymous

oops XD

- anonymous

this is tricky

- zepdrix

No no no :D we don't want multiple `differentials`. We just want 1 du!
And remember that you're `replacing (e^x dx) with du.
So that dx shouldn't be there anymore :O

- zepdrix

This is what we have so far,\[\Large \int\limits\limits\frac{e^x}{ (1+e ^{x}) ^{1/3}}\left(e^x\;dx\right) \qquad\to\qquad \int\limits\limits\frac{e^x}{ (u) ^{1/3}}\left(du\right)\]

- zepdrix

\[\Large \int\limits\limits\limits\frac{\color{#CC0033}{e^x}}{ (u) ^{1/3}}\left(du\right)\]

- zepdrix

Let's think back to our substitution that we set up,\[\Large u=1+\color{#CC0033}{e^x}\]

- anonymous

i see
wait let me continue the next step

- anonymous

\[\int\limits \frac{u+1 }{ u ^{1/3} } du\]

- anonymous

should be nagative i think

- zepdrix

oh on top?

- anonymous

u-1 :D

- zepdrix

\[\Large \int\limits\limits\limits\limits\frac{u-1}{u^{1/3}}\left(du\right)\]
Ok good good good :)

- zepdrix

From there we can split it into a couple of fractions.
Remember how to do that? :o

- zepdrix

With a problem like this, after you make your u-sub and get everything plugged in, it might not `look` simpler, but it is!
We have our addition/subtraction in the `numerator` now.
So we can take advantage of fraction math :O

- anonymous

yes \[\int\limits \frac{ u }{u ^{1/3}}-\frac{ 1 }{ u ^{1/3} } du\]

- zepdrix

cool, what next? :3

- anonymous

\[\int\limits \frac{ u }{ u ^{1/3} } - \int\limits \frac{ 1 }{u ^{1/3} } du\]

- zepdrix

sure, if you wanna write it that way, that's fine

- zepdrix

divide your u's in the first int :O

- anonymous

um theres an exponent on denominator u

- anonymous

will subtraction of exponent work?

- zepdrix

ya that sounds right.

- anonymous

\[\int\limits u ^{2/3} - \int\limits \frac{ 1 }{u ^{1/3} } du\]

- anonymous

what do i do with \[\int\limits \frac{ 1 }{ u ^{1/3} }\]

- zepdrix

\[\Large \frac{1}{x^2}\quad=\quad x^{-2}\]

- anonymous

haha then \[\int\limits u ^{-1/3}du do?\]

- anonymous

How did you get do? O_o .

- zepdrix

do? :o

- anonymous

wrote that @ wrong box :D

- zepdrix

oh haha

- anonymous

Alright. It's right then :P .

- anonymous

cool thanks now i just need to substitute u and make du a +c right?

- zepdrix

\[\Large \int\limits u^{2/3}-u^{-1/3}du\]Ya looks good so far :)

- zepdrix

Woops we still need to integrate !:O

- anonymous

Well I woudn;t substitute yet.

- anonymous

I would integrate it first and THEN substitute :P . NEVER substitute until the end haha.

- anonymous

\[\int\limits_{?}^{?} \frac{ u ^{5/3} }{5/3 } - \int\limits_{?}^{?}\frac{ u ^{2/3} }{ 2/3 } du\]

- anonymous

Well since you integrated the equation you can get rid of the intergal sign :P . There isn't any need of them since you solved the intergal.

- anonymous

But yep that's fine :) .

- anonymous

Now just simplify the thing :) .

- anonymous

ok cool i got the answer tnx everyone :0

- anonymous

Don't forget to add a +C!!

- anonymous

btw how did u know again that u = 1+e^x and not the other one,

- zepdrix

Hmm that's a good question.
Practice a little bit.
If you're ever unsure, always try the denominator first D:

- anonymous

Well you tend to pick u as something that it's derivative is also inside the intergal somewhere.

- anonymous

But ultimately it comes down to practice.

- anonymous

if i use u = e^x will it be ok? since i see there will be same du.

- anonymous

How would you deal with the e^x +1 in the demonitor then?

- zepdrix

If you let u=e^x you would have ended up with something like this,
\[\Large \int\limits \frac{u}{(1+u)^{1/3}}du\]

- zepdrix

Which unfortuately requires another substitution to solve easily.

- anonymous

And personally fraction exponents are quite hard to integrate :3 .

- anonymous

Often times in integrals you might have to use substitution more than once :P .

- anonymous

like integration by parts?

- anonymous

Maybe :P . It depends on the intergal.

- anonymous

Intergation is basically an experiment.

- anonymous

There may be multiple answers as well! :P .

- anonymous

cool can i ask another topic of integral(trigo sub) here? or should i close this and post a new one?

- anonymous

GO ahead :P .

- zepdrix

close it uppppp, it's getting long.
Too much code, makes it laggy :(

- anonymous

guess so :D

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