Question

integrate, using the substitution u=e^2x +4, {2/(e^2x +4)

Answer 1

\[\int\limits_{.}^{.} \frac{ 2 }{ e ^{2x} +4} dx \]

Answer 2

\[u=e ^{2x}+4\]

Answer 3

substitution of u gives...
\[\int\limits_{.}^{.}\frac{ 2 }{ u } dx\]

and \[dx=\frac{ du }{ 2e ^{2x} }\]

Answer 4

therefore
\[\int\limits_{.}^{.}\frac{ 2 }{ u } \frac{ du }{ 2e ^{2x} }\]

Answer 5

the twos cancel so...

Answer 6

\[\int\limits_{.}^{.} \frac{ 1 }{ ue ^{2x} } du\]

Answer 7

...but then I'm stuck, can you integrate ergardless of the fact there's still an x within th equation or do you have to substitute x for u?

Answer 8

e^2x = u - 4

Answer 9

IOW, you must switch it all to u...

Answer 10

oh right, I see

Answer 11

so you integrate:
\[\int\limits_{.}^{.} \frac{ 1 }{ u(u-4) } du\]
which is equal to~:
\[\int\limits_{.}^{.}\frac{ 1 }{ u ^{2}-4u }\]

Answer 12

Oh I was going to correct a mistake, but it looks like you've got it set up correctly! Hehe.
So yah. Partial Fraction Decomposition from there I guess.

Answer 13

partial what what??

Answer 14

\[\Large\rm \frac{1}{u(u-4)}=\frac{A}{u}+\frac{B}{u-4}\]Never learned this? :o

Answer 15

oh yeah i recognise that, so would A be (u-4) and B (u)?

Answer 16

No no, A and B end up being just numbers.

Answer 17

To figure out A and B, we'll multiply both sides through by that denominator on the left.
Giving us,\[\Large\rm 1=A(u-4)+Bu\]Ok with that step? :o

Answer 18

sorry yes that right

Answer 19

A= -1/4 and B = 5/4?

Answer 20

Hmm I think B=1/4... checking..

Answer 21

yeah sorry it is, i just did that quickly

Answer 22

\[\Large\rm \int\limits \frac{1}{u(u-4)}du=\int\limits \frac{A}{u}+\frac{B}{u-4}~du\]

\[\Large\rm =\int\limits \frac{-1/4}{u}+\frac{1/4}{u-4}~du\]So we get something like that, yah? :o

Answer 23

yep

Answer 24

\[\int\limits_{.}^{.} \frac{ -4 }{ u } + \frac{ 4 }{ u-4 }\]

Answer 25

No no no silly :3
We only flip the fraction when it's in the bottom.

Answer 26

\[\Large\rm =\int\limits\limits \frac{\color{orangered}{-1/4}}{u}+\frac{\color{orangered}{1/4}}{u-4}~du\]These are in the top :o

Answer 27

okayyy

Answer 28

You could do this I suppose, if you want it to be easier to read,
factor a 1/4 out of each term,\[\Large\rm =\frac{1}{4}\int\limits -\frac{1}{u}+\frac{1}{u-4}~du\]

Answer 29

ah okayy yeah

Answer 30

Remember how to integrate those two guys? :D

Answer 31

1/4 [ -lnu + ln Iu-4I ] ?

Answer 32

Mmm looks good.
Now to undo our sub :)

Answer 33

-lnu/4 + (lnIu-4I)/4 ?

Answer 34

We COULD distribute the 1/4 back in, but let's leave him out front for now maybe..
I think he's just gonna mess things up for us XD

Answer 35

okayy, sorry i misunderstood you before haha, so substitute u for e^2x +4 ?

Answer 36

\[\Large\rm \frac{1}{4}\left[\ln(\color{orangered}{u-4})-\ln(\color{orangered}{u})\right]\]Yah :) gotta do somethign with that orange stuff.

Answer 37

u-4 = e^2x
u = e^2x +4

Answer 38

\[\Large\rm \frac{1}{4}\left[\ln(\color{orangered}{e^{2x}})-\ln(\color{orangered}{e^{2x}+4})\right]\]Sounds good so far :D

Answer 39

1/4 [ 2x - ln(e^2x +4 ]

Answer 40

The first one simplifies? Ahh nice :D

Answer 41

You can throw the 1/4 in now if you want I suppose :P
But whatever, that's a fine looking answer.
Make sure to throw a +c somewhere in the mix since it's an `indefinite` integral.

Answer 42

Yayyy good job katie \c:/

Answer 43

okay thank youu, I'm soo bad at integration :( everything else I'm absolutely fine at but I missed a few lessons of integration and I'm just awful at it :L

Answer 44

There are so many little tricks and tools for integration :O
Takes a while to get comfortable with them.
Do tons and tons of practice problems to get a feel for what approach you want to take.

Answer 45

Like whenever you see a square in a denominator, or under a root,
a lightbulb should go off in your head saying "oh probably a trig sub"

Answer 46

A square with addition/subtraction*

Answer 47

yeah I certainly need too! The text book I have is awful though!

Answer 48

Are there any good websites to help me or anything?

Answer 49

Hmm there's probably a lot of cool stuff out there :d
I found this website to be really helpful when going through Differential Calculus:
https://www.math.ucdavis.edu/~kouba/ProblemsList.html
I think they have some decent practice problems for integration also.

What else is good? Uhhh
KhanAcademy,
tons of stuff on youtube. I found MIT's Calculus series really helpful.

Blah I dunno >.<

Answer 50

okay thank you, you've been really helpful!