Integrate ln(4+y^2)dy?

by dave_western
Tags: integrate, y2dy
dave_western is offline
Jan16-07, 04:05 PM
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Stuck at the end of a double integral, still have to integrate ln(4+y^2)dy

Assuming I did the right first step. Original double integral is


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benorin is offline
Jan16-07, 09:21 PM
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What are the bounds on x and y?
dave_western is offline
Jan16-07, 09:27 PM
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the region R = [1,2] * [0,1]

Hurkyl is offline
Jan16-07, 09:39 PM
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Integrate ln(4+y^2)dy?

Have you had any thoughts on integrating that? I see two obvious things to try:

(1) Do what you normally do with integrals of logarithms.
(2) Make a substitution.
mjsd is offline
Jan16-07, 10:22 PM
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eg. Integration by parts then trig sub.
dextercioby is offline
Jan17-07, 01:06 AM
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There's no need for substitution for that integral. Part integration once then a smart move in the numerator of the remaining integral and it's done.

murshid_islam is offline
Jan17-07, 03:39 AM
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is the integral we are talking about [tex]\int_{0}^{1}\int_{1}^{2}\frac{x}{x^2+y^2}dxdy[/tex]?

i get [tex]\frac{1}{2}\int_{0}^{1}\left[\ln(4+y^2)-\ln(1+y^2)\right]dy[/tex]
what should i do next? (edit: i got it. integration by parts.)
Gib Z
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Jan17-07, 06:34 PM
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Murshid, split the integral up, do them separately perhaps? Eg, say for [tex]\int ln(1+y^2) dy[/tex] we could let y^2 equal u. Find du/dy, easy. Then, solve for dy. Now substitute that value in. We end up with [tex]\int \frac{ln (u+1)}{2(u)^{1/2}} du [/tex]. Then some nice integration by parts and we are done?

Takes a while though, I hope your patient.
murshid_islam is offline
Jan18-07, 01:56 AM
P: 361
well i got it already. thanks anyway.
but we can directly use integration by parts on this [tex]\int ln(1+y^2) dy[/tex] by letting [tex]u = \ln(1+y^2)[/tex] and [tex]dv = dy[/tex]

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