Mastering the e^x Integral: Tips and Tricks from Eiano

[Eiano]

Hey, everyone

I am working on a calc problem, and I have no idea where to start. The integral is

e^x
------------- [division problem]
(25+e^2x)^4

Do I let my u equal to the 25+e^2x? or what...
then after that what do i do.

Thanks for all the help in advance.
-Eiano

 

[GCT]

Do I let my u equal to the 25+e^2x? or what...
then after that what do i do.

here's an ingenious idea, try it out, yes try the substitution you mentioned and we'll see what magically happens.

 

[Eiano]

Anyone else?
(i KNOW my u can't be 25+e^2x, that's what th OR WHAT was for)
but HAHA, u were so funny with that response.. haha

 

[LeonhardEuler]

Partial fraction decomposition? It looks like it'll take a long time, though.

 

[whozum]

Anyone else?
(i KNOW my u can't be 25+e^2x, that's what th OR WHAT was for)
but HAHA, u were so funny with that response.. haha

He was serious.. that's the solution.

 

[GCT]

uhmm, actually I wasn't, he he

partial fractions...hmm...sounds interesting, we'll have to try it out, although I've never done partial fractions with e^x,e^2x as a variable.

$$u=(25+e^{2x}),~du=2e^{2x}dx,~e^{x}= \sqrt{u-25}$$

next substitute u=1/t

simplifying will give you soley a square root function in the denominator, convert this to a completed square, the rest should be easy

I'm sure there's a more elegant solution though

latex is ****ing up, I've actually typed the whole problem and solution out through latex, however for some reason it's picking up some old latex data from a couple of months ago

 

[GCT]

alright finally got the latex to appear

$$I = \int \frac{e^{x}dx}{(25+e^{2x})^4}$$

$$u=(25+e^{2x}),~du=2e^{2x}dx,~e^{x}= \sqrt{u-25}$$

$$I=.5 \int \frac{du}{u^{4} \sqrt{u-25}}$$

$$u=1/t,~du=-1/t^{2} dt$$

$$I=-.5 \int \frac{dt}{ \sqrt{1/t^{2} - 25/t}}$$

$$-.5 \int \frac{dt}{ \sqrt{(1/t -12.5)^{2} - 12.5^{2}}}$$

$$(1/t -12.5) = 12.5sec \theta$$

the rest should be easy

etc...

see any errors, please point them out, I'm guessing that there's probably a more elegant solution

 

[HallsofIvy]

But e^2x is (e^x)² so 25+ e^2x= 25+ (e^x)². Try letting u= e^x first, so that du= e^xdx. Now the integral is $$\int \frac{du}{25+u^2}$$ and that smells like an arctangent. Let u= 5 v so that du= 5 dv and 25+ u²= 25(1+ v²). Now the integral is $$\frac{5}{5}\int \frac{dv}{v^2+1}= arctan(v)+ C$$.
Of course, that's $$arctan(\frac{u}{5})+ C= arctan(\frac{e^x}{5})+ C$$

 

[TD]

Unfortunately, there was a 4th power in the denominator...

 

[GCT]

But e2x is (ex)2 so 25+ e2x= 25+ (ex)2. Try letting u= ex first, so that du= exdx. Now the integral is

I'm not quite sure what you're getting at there can you clarify? also note as TD said, there's a fourth power in the denominator.

I'm not able to see anything wrong with my solution.

 

[GCT]

HallsofIvy, nevermind, I thought you were pointing out an error in my solution

 

[TD]

If it weren't for the fourth power, HallsofIvy would've had an easy solution. He just must have missed it, I don't think he meant that yours was wrong. Without the fourth power, yours just seemed so long

 

[GCT]

his method might lead to a partial fractions solution

 

[LeonhardEuler]

$$I=.5 \int \frac{du}{u^{4} \sqrt{u-25}}$$
$$u=1/t,~du=-1/t^{2} dt$$
$$I=-.5 \int \frac{dt}{ \sqrt{1/t^{2} - 25/t}}$$

I think I might have found an error here, GCT. Substituting gives:
$$I=.5 \int \frac{(-1/t^2)dt}{ (1/t)^4\sqrt{1/t - 25}}$$
$$=-.5 \int \frac{dt}{ (1/t)^2\sqrt{1/t - 25}}$$
$$=-.5 \int \frac{dt}{ \sqrt{1/t^5 - 25/t^4}}$$

 

[GCT]

I'm not quite sure what you did there, note that bringing in 1/t^2 within the square root will reduce it to 1/t within the square root.

 

[LeonhardEuler]

Doesn't it go like this:
$$\sqrt{a}\sqrt{b}=\sqrt{ab}$$
$$\frac{1} {t^2}\sqrt{x} = \sqrt{\frac{1} {t^4}}\sqrt{x}=\sqrt{\frac{x} {t^4}}$$

 

[GCT]

yeah, you're right

 

[GCT]

alright, shame on me

What we can do is

$$I=.5 \int \frac{du}{u^{4} \sqrt{u-25}}$$
$$u=25sec^{2} \theta ,~du=25tan \theta d \theta$$
$$I= \frac{-1}{10(25^{3})} \int cos^{8} \theta d \theta$$
which can be solved using standard procedure ("table integral")

hopefull I haven't goofed up this time, I'll be posting the full version later

 

[LeonhardEuler]

Oh, man bad news, GCT. If $u=\sec^2{\theta}$ then
$$\frac{du}{d\theta}=\tan{\theta}\sec^2{\theta}$$
you were probably thinking of
$$\int \sec^2{\theta}=\tan{\theta}$$
It's Ok, we all have bad days!

 

[GCT]

I think I've been drinking way too much these days, anyways thanks for pointing that out.

So the modification would result in
$$I= \frac{-1}{10(25^{3})} \int cos^{6} \theta d \theta$$
...right?

 

[GCT]

so, so far I have

$$I = \int \frac{e^{x}dx}{(25+e^{2x})^4}$$


$$u=(25+e^{2x}),~du=2e^{2x}dx,~e^{x}= \sqrt{u-25}$$

$$I=.5 \int \frac{du}{u^{4} \sqrt{u-25}}$$

$$u=25sec^{2} \theta ,~du=50tan \theta sec^{2}\theta d \theta$$

$$\frac{-1}{25^{2}10} \int cos^{6} \theta d \theta$$

which can be solved easily using standard procedure (trignometric integrals, formula should be listed within the list of table integral-back of text)

 

[LeonhardEuler]

Looks right to me.