Hey guys, so I'm taking differential equations and I feel

>>8322417
just integrate it by parts, family

The integral of [math]1/f(x) = f(x)^{-1}[/math] is [math]0[/math].

>>8322417
[math]\int\frac{1}{e^{2x}+4}dx=\frac{1}{4}\int\frac{1}{(\frac{e^x}{2})^2+1}dx[/math]

Which will be an arctan(u) integral...

>>8322417
Just eye balling it. By parts you set u = e^2x + 4 so du = 2e^2x
dv = dx
v = x

vdu = 2xe^2x
Use u substitution here

Holy shit thanks so much man I can't believe I didn't recognize that arctan property!

Allright:
[math]=0.25(\int\frac{exp(2x)+4}{exp(2x)+4}dx-2\int\frac{0.5exp(2x)}{exp(2x)+4}dx)[/math]
If you're not a complete retard you can solve this.

>>8322433
Wrong e^x =/= x

>>8322456
Yea I realized something was wrong when I started doing it.

>>8322417
1/(e^2+4)
That's an Arc Tan.
1/u^2+v^2

Should get you (1/4)Arctan(e/2) I think. It's been a while since I did inverse trig int

>>8322452
How did you come up with this?

partial fraction decomposition

>>8322417
https://www.symbolab.com/solver/equation-calculator/%5Cint%5Cfrac%7B1%7D%7Be%5E%7B2x%7D%2B4%7Ddx

Don't use this as a crutch

>>8322463
No, you.

>>8322456

>>8322474
FUCK

u sub then Arctan?

Well guys I may just end up killing myself

>>8322417
Use use a u substitution and use partial fractions, x = (1/2)*ln(u) will get you the right answer.
If you want to solve this with trial and error, you can recall that differentiating ln(e^2x + 4) will get you ((e^2x)*2)/(e^2x + 4). In order to cancel the numerator and denominator, we can add 0, which gets us ((e^2x)*2 + 4*2)/(e^2x + 4) - 4*2/(e^2x + 4), which is equal to 2 - 8/(e^2x + 4). Now, scaling ln(e^2x + 4) down by (-1/8) will change our derivative to 1/(e^2x + 4) - 1/4, so, to get rid of the -1/4, we need to add the antiderivative of 1/4 to (-1/8)*ln(e^2x + 4), which is x/4. So the answer is x/4 - (1/8)*ln(e^2x + 4) + C.

>>8322464
It's intuition combined with practice I think. Been so long since I've studied this shit. It's just one of those well known methods where you split up the integral in two easier ones.

Basically you try to match the numerator to the denomerator. Because I see 4 in the denomerator, I multiply the numerator by 4, but then you also have to divide the integral by 4 to keep the same result:
[math]=0.25\int\frac{4}{exp(2x)+4}dx[/math]
Then I see the exp(2x) in the denomerator, so I add exp(2x) to the numerator. Then I also have to substract it for obvious reasons, the result you can split up:
[math]=0.25(\int\frac{exp(2x)+4}{exp(2x)+4}dx-\int\frac{exp(2x)}{exp(2x)+4}dx)[/math]
Last step is fairly obvious and you get>>8322452

Which is two easy to solve integrals.

>>8322456
Nigger are you retarded? Use u-substitution. Or go back to your Calc I class you fucking peasant.

>>8322500
It's you who is the retard. Try the substitution and you'll see your wrong. I already posted an easy solution see>>8322489

Fuck Calc I, i never took a shitty calc class. Real analysis is much better.

Go back to Calc I dumbo haha

>>8322433
no it isn't, when you do the u substitution to get rid of e^x, you end up with 1/[(u)*(u^2 + 1)]. Partial fraction decomposition turns this into 1/u - u/(u^2 + 1). Integrating these simpler terms, you get ln(u) - (1/2)*ln(u^2 + 1) + C. Factoring back in the 1/4 you took out and putting u = (1/2)*e^x back in, we get x/4 - (1/4)*ln(2) - (1/8)*ln(((1/2)*e^x)^2 + 1) + C, which, taking the logarithm identities and C into account, is just x/4 - (1/8)*ln(e^2x + 4) + C.
C.

>>8322489
Thanks so much man I really feel like shit that I wasn't able to do this on my own.

>>8322551
Analysis problems like that are nothing but pure practice. There isn't a general method, just gotta be used enough to them that you "see" the trick.