anonymous
  • anonymous
Integral Calculus (Indefinite) what pointers should i think to choose the correct u. in u-substitution
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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zepdrix
  • 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
  • zepdrix
In the example:\[\Large u=x^3+4 \qquad\qquad\qquad du=\left(3x^2\;dx\right)\]
zepdrix
  • 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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zepdrix
  • 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
  • zepdrix
So in this example:\[\Large u=\ln x \qquad\qquad\qquad du=\left(\frac{1}{x}dx\right)\]
zepdrix
  • 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
  • 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
  • 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
  • zepdrix
Do you have any specific examples you need help with? :o Or any clarification on what I did above?
dumbcow
  • dumbcow
@zepdrix , very thorough examples :)
anonymous
  • 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
  • zepdrix
For a problem like this?\[\Large \int\limits e^{3x}dx\]Or something a bit more complex?
anonymous
  • anonymous
something like this \[\int\limits \frac{ e ^{2x} }{ (1+e ^{x}) ^{1/3}} dx\]
anonymous
  • anonymous
u see there are 2 \[e ^{x}\] but the other one is \[e ^{2x}\] how do i deal with this?
zepdrix
  • zepdrix
mm sec XD thinking hehe
zepdrix
  • 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
  • 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
  • anonymous
ok im at it, ill do my best
zepdrix
  • zepdrix
\[\Large \int\limits\frac{e^x}{ (1+e ^{x}) ^{1/3}}\left(e^x\;dx\right)\]
zepdrix
  • zepdrix
So I split up the square which allowed me to group it a little weird.
anonymous
  • anonymous
haha never thought of doin that
anonymous
  • anonymous
but what sould i use as a U
zepdrix
  • 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
  • zepdrix
So what do you get for du? :x
anonymous
  • anonymous
e^x
zepdrix
  • 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
  • anonymous
\[\int\limits \frac{(du)(du) }{(u)^{1/3} } dx\]
anonymous
  • anonymous
oops XD
anonymous
  • anonymous
this is tricky
zepdrix
  • 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
  • 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
  • zepdrix
\[\Large \int\limits\limits\limits\frac{\color{#CC0033}{e^x}}{ (u) ^{1/3}}\left(du\right)\]
zepdrix
  • zepdrix
Let's think back to our substitution that we set up,\[\Large u=1+\color{#CC0033}{e^x}\]
anonymous
  • anonymous
i see wait let me continue the next step
anonymous
  • anonymous
\[\int\limits \frac{u+1 }{ u ^{1/3} } du\]
anonymous
  • anonymous
should be nagative i think
zepdrix
  • zepdrix
oh on top?
anonymous
  • anonymous
u-1 :D
zepdrix
  • zepdrix
\[\Large \int\limits\limits\limits\limits\frac{u-1}{u^{1/3}}\left(du\right)\] Ok good good good :)
zepdrix
  • zepdrix
From there we can split it into a couple of fractions. Remember how to do that? :o
zepdrix
  • 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
  • anonymous
yes \[\int\limits \frac{ u }{u ^{1/3}}-\frac{ 1 }{ u ^{1/3} } du\]
zepdrix
  • zepdrix
cool, what next? :3
anonymous
  • anonymous
\[\int\limits \frac{ u }{ u ^{1/3} } - \int\limits \frac{ 1 }{u ^{1/3} } du\]
zepdrix
  • zepdrix
sure, if you wanna write it that way, that's fine
zepdrix
  • zepdrix
divide your u's in the first int :O
anonymous
  • anonymous
um theres an exponent on denominator u
anonymous
  • anonymous
will subtraction of exponent work?
zepdrix
  • zepdrix
ya that sounds right.
anonymous
  • anonymous
\[\int\limits u ^{2/3} - \int\limits \frac{ 1 }{u ^{1/3} } du\]
anonymous
  • anonymous
what do i do with \[\int\limits \frac{ 1 }{ u ^{1/3} }\]
zepdrix
  • zepdrix
\[\Large \frac{1}{x^2}\quad=\quad x^{-2}\]
anonymous
  • anonymous
haha then \[\int\limits u ^{-1/3}du do?\]
anonymous
  • anonymous
How did you get do? O_o .
zepdrix
  • zepdrix
do? :o
anonymous
  • anonymous
wrote that @ wrong box :D
zepdrix
  • zepdrix
oh haha
anonymous
  • anonymous
Alright. It's right then :P .
anonymous
  • anonymous
cool thanks now i just need to substitute u and make du a +c right?
zepdrix
  • zepdrix
\[\Large \int\limits u^{2/3}-u^{-1/3}du\]Ya looks good so far :)
zepdrix
  • zepdrix
Woops we still need to integrate !:O
anonymous
  • anonymous
Well I woudn;t substitute yet.
anonymous
  • anonymous
I would integrate it first and THEN substitute :P . NEVER substitute until the end haha.
anonymous
  • anonymous
\[\int\limits_{?}^{?} \frac{ u ^{5/3} }{5/3 } - \int\limits_{?}^{?}\frac{ u ^{2/3} }{ 2/3 } du\]
anonymous
  • 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
  • anonymous
But yep that's fine :) .
anonymous
  • anonymous
Now just simplify the thing :) .
anonymous
  • anonymous
ok cool i got the answer tnx everyone :0
anonymous
  • anonymous
Don't forget to add a +C!!
anonymous
  • anonymous
btw how did u know again that u = 1+e^x and not the other one,
zepdrix
  • zepdrix
Hmm that's a good question. Practice a little bit. If you're ever unsure, always try the denominator first D:
anonymous
  • anonymous
Well you tend to pick u as something that it's derivative is also inside the intergal somewhere.
anonymous
  • anonymous
But ultimately it comes down to practice.
anonymous
  • anonymous
if i use u = e^x will it be ok? since i see there will be same du.
anonymous
  • anonymous
How would you deal with the e^x +1 in the demonitor then?
zepdrix
  • 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
  • zepdrix
Which unfortuately requires another substitution to solve easily.
anonymous
  • anonymous
And personally fraction exponents are quite hard to integrate :3 .
anonymous
  • anonymous
Often times in integrals you might have to use substitution more than once :P .
anonymous
  • anonymous
like integration by parts?
anonymous
  • anonymous
Maybe :P . It depends on the intergal.
anonymous
  • anonymous
Intergation is basically an experiment.
anonymous
  • anonymous
There may be multiple answers as well! :P .
anonymous
  • anonymous
cool can i ask another topic of integral(trigo sub) here? or should i close this and post a new one?
anonymous
  • anonymous
GO ahead :P .
zepdrix
  • zepdrix
close it uppppp, it's getting long. Too much code, makes it laggy :(
anonymous
  • anonymous
guess so :D

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