Question

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

in u-substitution

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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.\[\Large \int\limits\ln x\left(\frac{1}{x}dx\right)\]

Answer 5

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

Answer 6

It's not always that straight forward though.
Sometimes you need to apply a `u-sub` and then mess around with the pieces.

Example:\[\Large \int\limits \frac{x}{\sqrt{x-2}}dx\]

Answer 7

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.\[\Large u=x-2 \qquad\to\qquad x=u+2\]

Answer 8

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

Answer 9

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

Answer 10

@zepdrix , very thorough examples :)

Answer 11

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?

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

mm sec XD thinking hehe

Answer 16

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

Answer 17

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.

Answer 18

ok im at it, ill do my best

Answer 19

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

Answer 20

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

Answer 21

haha never thought of doin that

Answer 22

but what sould i use as a U

Answer 23

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 :)


\(\Large u=1+e^x\)

Answer 24

So what do you get for du? :x

Answer 25

e^x

Answer 26

\[\Large du=(e^x\;dx)\]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

Answer 27

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

Answer 28

oops XD

Answer 29

this is tricky

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

i see
wait let me continue the next step

Answer 35

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

Answer 36

should be nagative i think

Answer 37

oh on top?

Answer 38

u-1 :D

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

cool, what next? :3

Answer 44

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

Answer 45

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

Answer 46

divide your u's in the first int :O

Answer 47

um theres an exponent on denominator u

Answer 48

will subtraction of exponent work?

Answer 49

ya that sounds right.

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

How did you get do? O_o .

Answer 55

do? :o

Answer 56

wrote that @ wrong box :D

Answer 57

oh haha

Answer 58

Alright. It's right then :P .

Answer 59

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

Answer 60

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

Answer 61

Woops we still need to integrate !:O

Answer 62

Well I woudn;t substitute yet.

Answer 63

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

Answer 64

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

Answer 65

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.

Answer 66

But yep that's fine :) .

Answer 67

Now just simplify the thing :) .

Answer 68

ok cool i got the answer tnx everyone :0

Answer 69

Don't forget to add a +C!!

Answer 70

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

Answer 71

Hmm that's a good question.
Practice a little bit.

If you're ever unsure, always try the denominator first D:

Answer 72

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

Answer 73

But ultimately it comes down to practice.

Answer 74

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

Answer 75

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

Answer 76

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

Answer 77

Which unfortuately requires another substitution to solve easily.

Answer 78

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

Answer 79

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

Answer 80

like integration by parts?

Answer 81

Maybe :P . It depends on the intergal.

Answer 82

Intergation is basically an experiment.

Answer 83

There may be multiple answers as well! :P .

Answer 84

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

Answer 85

GO ahead :P .

Answer 86

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

Answer 87

guess so :D