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not sure how to do this one......seems i get ln on both sides....but dont know where to go from there.

And if you could, could you also explain how to generally solve for y if in fact it is in ln format? - for ex...... ln(y) =.............

2007-08-12 17:14:04 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

4 answers

∫ (1/y) dy = ∫ [x / (x² + 1) ] dx
ln y = (1/2) ln (x² + 1) + C
2 ln y = ln (x² + 1) + 2C
2 ln y = ln (x² + 1) + ln K
y ² = K (x² + 1)

2007-08-12 22:03:03 · answer #1 · answered by Como 7 · 1 0

After a bit or rewriting, and then a bit of integration, I also get ln on both sides, but what's wrong with that? If you're solving for y and you have:

lny = (whatever), then remember that ln is really log base e, so rewriting what I have above will be:

y = e ^ (whatever)

It sounds like you're doing the right things, though! Good luck!

2007-08-13 00:42:06 · answer #2 · answered by douglas 2 · 0 0

Separate variables,
dy/y = x/(x^2+1) dx
Integrate both sides,
lny = (1/2)ln(x^2+1)+c'
y = e^[ (1/2)ln(x^2+1)+c']
or
y = c√(x^2+1), where c = e^c' is a constant

2007-08-13 00:48:52 · answer #3 · answered by sahsjing 7 · 1 0

Good luck is all I have to say. Oh hi!

2007-08-13 00:23:25 · answer #4 · answered by ? 6 · 0 0

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