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Topic: integrals
Replies: 2   Last Post: Feb 17, 2013 1:18 PM

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johnykeets

Posts: 8
From: usa
Registered: 1/1/13
Re: integrals
Posted: Jan 22, 2013 3:56 AM
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First let us take the problem,
Here, 1/x is taken as the input for the given function.
It's the difference between f(x)*g(x) and f(g(x)), with this case being the latter. As such, there is no way to separate the 1/x from the arctangent -- they MUST be considered as a single unit.
Then, we let u=arctan (1/x), du = 1/(1+(1/x)²) * (-1/x²) dx = -1/(x²+1) dx (using the chain rule), dv = dx, v = x. Then we have:

[1, ?3]?arctan (1/x) dx
x arctan (1/x) |[1, ?3] + [1, ?3]?x/(x²+1) dx
?3 arctan (1/?3) - 1 arctan 1 + [1, ?3]?x/(x²+1) dx
??3/6 - ?/4 + [1, ?3]?x/(x²+1) dx

Now making the substitution u=x²+1, du=2x dx, x dx = du/2, x=1 ? u=2, x=?3 ? u=4, we have:

??3/6 - ?/4 + 1/2 [2, 4]?1/u du
??3/6 - ?/4 + 1/2 ln |u| |[2, 4]
??3/6 - ?/4 + 1/2 (ln 4 - ln 2)
??3/6 - ?/4 + 1/2 ln (4/2)
??3/6 - ?/4 + ln (?2)
Replace (22/7, the value of Pie in place of ?)

Hence Proved.


Message was edited by: johnykeets


Message was edited by: johnykeets


Date Subject Author
1/19/13
Read integrals
ungeheuer
1/23/13
Read Re: integrals
johnykeets
2/17/13
Read Re: integrals
Fabien

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