# Help with e^x intgral

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

Homework Helper
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

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

whozum
Eiano said:
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.

Homework Helper
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

Homework Helper
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

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Homework Helper
But e2x is (ex)2 so 25+ e2x= 25+ (ex)2. Try letting u= ex first, so that du= exdx. 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 $$arctan(\frac{u}{5})+ C= arctan(\frac{e^x}{5})+ C$$

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

Homework Helper
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.

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

Homework Helper
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

Homework Helper
his method might lead to a partial fractions solution

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

Homework Helper
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.

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

Homework Helper
yeah, you're right

Homework Helper
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

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Gold Member
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!

Homework Helper
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?

Homework Helper
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)

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Gold Member
Looks right to me.