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# Integration by Substitution Watch

1. I have been given a question and I have tried multiple methods and I still can't get the answer of '1/6ln2' or '1/ln4 - 1/ln8'.

Use the substitution u=2^x to find the exact value of

2^x/(2^x+1)^2 dx. Limits of x are 1 and 0.

I have tried all kinds of things, and I'd very much appreciate some form of way doing this.
2. (Original post by JLangley753)
I have been given a question and I have tried multiple methods and I still can't get the answer of '1/6ln2' or '1/ln4 - 1/ln8'.

Use the substitution u=2^x to find the exact value of

2^x/(2^x+1)^2 dx. Limits of x are 1 and 0.

I have tried all kinds of things, and I'd very much appreciate some form of way doing this.
Hint:

Use this fact when making the substitution.
3. (Original post by Indeterminate)
Hint:

Use this fact when making the substitution.
I'm at the stage where I've got 1/(u+1)^2 du with limits 2 and 1.

Is this where I'd input this into the substitution?
4. (Original post by JLangley753)
I'm at the stage where I've got 1/(u+1)^2 du with limits 2 and 1.

Is this where I'd input this into the substitution?
You get

Now, you only have a 2 to the x on the top. So if you multiply this by ln 2 to substitute the above, you need to divide by ln 2 too to AVOID changing the integral.
5. (Original post by Indeterminate)
You get

Now, you only have a 2 to the x on the top. So if you multiply this by ln 2 to substitute the above, you need to divide by ln 2 too to AVOID changing the integral.
Surely then I'm left with the equivalent u^2/(u+1)^2 but with ln2 on the top and bottom?

I hope this comes to me eventually.
6. (Original post by JLangley753)
Surely then I'm left with the equivalent u^2/(u+1)^2 but with ln2 on the top and bottom?

I hope this comes to me eventually.
No.

You want to replace the RHS with the LHS in the integral, right?

Therefore, you have to multiply it ln 2 and then that numerator all becomes du.

Now, you've changed the integral by multiplying by ln 2, so you have to divide the integral that's with respect to u i.e the one with du in it, by ln 2 as well
7. (Original post by JLangley753)
I have been given a question and I have tried multiple methods and I still can't get the answer of '1/6ln2' or '1/ln4 - 1/ln8'.

Use the substitution u=2^x to find the exact value of

2^x/(2^x+1)^2 dx. Limits of x are 1 and 0.

I have tried all kinds of things, and I'd very much appreciate some form of way doing this.

gives

So

so

8. (Original post by TenOfThem)

gives

So

so

I've got this, absolutely fine. I know the integral of the first part is 1/u+1 + ln(u+1) but do I need to integrate 1/uln2? I have never integrated something like that before.
9. 1/ln2 is a constant...

EDIT: and where does integrating 1/uln2 come from? You only need to integrate 1/(u+1)^2 (times 1/ln2)
10. (Original post by JLangley753)
I've got this, absolutely fine. I know the integral of the first part is 1/u+1 + ln(u+1) but do I need to integrate 1/uln2? I have never integrated something like that before.
erm, no

you need to cancel the u in the numerator with the u in the denominator

then integrate
11. Because you say "integral of the first part", perhaps your error is that you're treating the multiplication sign as an addition sign?
12. (Original post by TenOfThem)
erm, no

you need to cancel the u in the numerator with the u in the denominator

then integrate
Argh, thank you. Sorry, I actually don't know how I didn't get this before. I should have realised what Indeterminate was talking about in his first post.

From here I took the ln2 outside, integrated to get -(u+1)^-1 and the rest is a treat. Input the limits to get 1/6, multiply that by ln2 to get 1/6ln2.

Thank you guys, seriously. I apologise for being so slow at times.
13. (Original post by JLangley753)
Argh, thank you. Sorry, I actually don't know how I didn't get this before. I should have realised what Indeterminate was talking about in his first post.

From here I took the ln2 outside, integrated to get -(u+1)^-1 and the rest is a treat. Input the limits to get 1/6, multiply that by ln2 to get 1/6ln2.

Thank you guys, seriously. I apologise for being so slow at times.

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Updated: March 23, 2013
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