Solve Integral: e^(x)/(e^(2X) + 9)

[foges]

Hey, just did a test and have no clue about this question:

Integrate:

e^(x)/(e^(2X) + 9)

Really curious how this is done. Thanx

 

[cristo]

Well, the numerator is (a constant times) the derivative of the denominator. Can you integrate a function of the form \frac{f'(x)}{f(x)}?

 

[foges]

I thought that too at first, but then realized that the numerator is e^(x) and not e^(2x), which would have made it a lot simpler. had it been e^(2x) i could have used ln(f(x))

 

[da_willem]

Do a change of variables e^x to z, this yields a standard integral of the arctangent.

 

[cristo]

Sorry, I misread that! Anyway, willem's idea seems the way to go!

 

[foges]

Yeah i tried that on the test, but didnt get too far, probably cause i havnt worked too much with arctangents (arcsins or arccosins for that matter) in calculus.

I prety much put down (1/3)arctan(e^(x)/3), but it gave me the wrong answer (PS: it was the finite integral from 0 to ln(3), so i could test it on my calculator)

 

[Gib Z]

Well \int \frac{e^x}{(e^x)^2 +9} dx is what makes the substitution u=e^x easier to see. du = e^x dx

That makes the integral \int \frac{1}{u^2+9} du which is of the arctan form, but if you haven't learned that then use the substitution u= 3 tan theta.

 

[foges]

Thanx,
Well my formula booklet says: integral of 1/(a^2+x^2) = (1/a)arctan(x/a) + c

Using that logic and substitution, i got: (1/3)arctan(e^(x)/3), which isn't right. what did i do wrong here?

Using e^x =3tan(theta) I am left with dx= 3sec^2(theta) d(theta), which turns out ugly

 

[Gib Z]

Forget about replacing u with e^x with respect to my suggested substitution for now.

\int \frac{1}{u^2+3^2} du = \frac{1}{3} \int \frac{1}{\sec^2 \theta} \cdot \sec^2 \theta d\theta = \frac{1}{3} \theta + C = \frac{1}{3} \arctan (u/3) + C = \frac{1}{3} \arctan (\frac{e^x}{3}) +C which turns out the same as your formula booklet! that's because for solving that general integral, we use u=a tan theta!

Why in the world do you think that's not correct!

 

[foges]

for two reasons, firstly because calculating the finite ingtegral from 0 to ln(3) i get a different answer than using that formula (TI-83 plus). Secondly, when graphing that equation and the original equation it doesn't look like the first derivative to me.

Maybe I am just screwed up and tried everything in the wrong mode or something. Using that formula, the answer will be in radians right?

Thanx for your time btw

 

[Gib Z]

When EVER we do calculus, everything is in Radians thanks :P

Now the answer I get is \pi/12 - 1/3 \arctan (1/3), which turns out around 0.154549203. If that TI 83 plus is different, then I can finally be assured i know more than a calculator.

 

[Office_Shredder]

foges, when you say taking the finite integral from 0 to ln(3), you mean you take the integral from 0 to ln(3) of the original integral in your first post, then compare it to

\frac{1}{3}*(artcan(e^{ln(3)}/3) - arctan(e^0/3))

 

[foges]

Ok, just re did it on my calculator and i got that answer, so strange, maybe i subtracted arctan(0) or had it in the wrong mode or something. Anyways, thanks for the help.