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# Integration by substitution Watch

1. Okay so I have a substitution question, and the answer I'm getting isn't the same as is in the textbook. This is the question:

Integrate 2/(e^2x + 4) dx, using the substitution u = e^2x + 4

Any help would be much appreciated
2. This is (e^2x) + (4) by the way, not e^(2x + 4)
3. (Original post by PeteyB26)
Okay so I have a substitution question, and the answer I'm getting isn't the same as is in the textbook. This is the question:

Integrate 2/(e^2x + 4) dx, using the substitution u = e^2x + 4

Any help would be much appreciated

Differentiate u,

Now sub the data in the integral.

Remember you will have to use partial fractions.
4. Yes I got that, can I not use the integrating fractions method? (int) f'(x)/f(x) = ln(fx) + c
5. (Original post by PeteyB26)
Yes I got that, can I not use the integrating fractions method? (int) f'(x)/f(x) = ln(fx) + c
What do you get as your integral?

I don't think the integral is in the form f'(x)/f(x), so you can't use this method, you need to use partial fraction.
6. For the integral I have 1/u(u-4)

But if I take out 1/2u-4 then I can change the integral to 2u-4/u(u-4) which would then be in the f'(x)/f(x) form?

Just because we've not done integration with partial fractions yet, so I would presume this question can be done without them..?
7. (Original post by PeteyB26)

But if I take out 1/2u-4
8. (Original post by PeteyB26)
For the integral I have 1/u(u-4)

But if I take out 1/2u-4 then I can change the integral to 2u-4/u(u-4) which would then be in the f'(x)/f(x) form?

Just because we've not done integration with partial fractions yet, so I would presume this question can be done without them..?
I have no idea what you are saying but I suspect it is terrible.

Did you think ?
9. Sorry, just done it with partial fractions and it worked, thank you!

Out of interest though, why would that method give me a different answer? Is it because I took a value with u out of the integral? Not sure if I can do that?
10. I think it can only be don by using partial fractions when you get to I= integral (1/u(u-4))

If it helps, (1/u(u-4)) = 1/4((1/u-4) - (1/u)) which can be integrated pretty easily
11. (Original post by PeteyB26)
For the integral I have 1/u(u-4)

But if I take out 1/2u-4 then I can change the integral to 2u-4/u(u-4) which would then be in the f'(x)/f(x) form?

Just because we've not done integration with partial fractions yet, so I would presume this question can be done without them..?
I don't understand the text in red, how can you take a function out?

You will probably know how to do partial fractions.

Just find the partial fractions first.
12. (Original post by PeteyB26)
Sorry, just done it with partial fractions and it worked, thank you!

Out of interest though, why would that method give me a different answer? Is it because I took a value with u out of the integral? Not sure if I can do that?
It should give you the same answer, the books are wrong sometimes
13. (Original post by Mr M)
I have no idea what you are saying but I suspect it is terrible.

Did you think ?
No, the differential of u(u-4) is 2u - 4

I was trying to get the integral into f'(x)/f(x) form so to convert the 1 into the differential of the denominator I took out 1/2u-4, this is where my error was... It's okay I see it now xD
14. (Original post by PeteyB26)
For the integral I have 1/u(u-4)

But if I take out 1/2u-4 then I can change the integral to 2u-4/u(u-4) which would then be in the f'(x)/f(x) form?
Remember, you can take out constants only.
15. (Original post by PeteyB26)
Sorry, just done it with partial fractions and it worked, thank you!

Out of interest though, why would that method give me a different answer? Is it because I took a value with u out of the integral? Not sure if I can do that?
When you say "take out"

Do you meant take out of the integral like you would "take out" a factor

If so

NO
16. If you wanted to avoid partial fractions you could have made a second substitution t = u-2 and then the integral is in a standard form you can look up in your formula book. Partial fractions is the easier way in my opinion though.
17. (Original post by Dog4444)
Remember, you can take out constants only.

(Original post by TenOfThem)
When you say "take out"

Do you meant take out of the integral like you would "take out" a factor

If so

NO

Yes, I see now, you can only take out constants. Thank you all! Partial fractions are the way forward! Suppose that teaches you not to do questions for which you've not yet been taught the content...
Potential for hours and hours of frustration haha!

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Updated: April 5, 2012
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