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integration by substitution question (1 Viewer)

shaon0

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Can somebody help me with this one.


using
I=S dx/(4+x^2)^3/2
Let x=2tan@, dx=2sec^2@ d@
I=S 2sec^2 @ d@ / 4^(3/2).sec ^3 @
=1/4 S cos@ d@
=1/4.sin@ +C
= 1/4.sin(atan(x/2)) +C
= 1/4.x/sqrt(4+x^2) +C
 
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nrlwinner

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Thanks. I got another one here. I've tried so hard but I can't seem to get the answer which is 7.



I've been using the substitution u^2 = x+1
 

shaon0

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Thanks. I got another one here. I've tried so hard but I can't seem to get the answer which is 7.



I've been using the substitution u^2 = x+1
I=S[3,8] (x+1-1)/sqrt(x+1)(sqrt(x+1)-1) dx
Let u^2=x+1, 2u du=dx:
I=S[2,3] (u^2-1)/u(u-1) 2u du
=2 S[2,3] (u+1) du
=2 [1/2u^2+u] {2,3}
=2 (9/2+3-2-2)
=2(1/2+3)
=7
 

nrlwinner

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Can someone show me where to go next. I've done a question up to here.



I've been using t=tanx/2 and I'm trying to prove for 1/4ln3. If it's impossible from here, please tell me.
 

Drongoski

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Not very straightforward. I'll skip the details.





OK . . . 1/2 my integral = 1/4 ln 3.
 
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nrlwinner

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Not very straightforward. I'll skip the details.





OK . . . 1/2 my integral = 1/4 ln 3.
Brilliant. But how did you know to divide the denominator like that because it is not reducible. (1st line I mean)
 

shaon0

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Brilliant. But how did you know to divide the denominator like that because it is not reducible. (1st line I mean)
Terms of odd powers cancel, so the middle term of the quadratic has to be minus in this case.
 

Drongoski

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Brilliant. But how did you know to divide the denominator like that because it is not reducible. (1st line I mean)
Yes! But not a soul on bos has hired me to tutor!
 
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Drongoski

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Re 1st line of my solution:

 
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