Partial fraction? or using properties?

shseo0315
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Homework Statement



dx/dt = 9-4x^2 , x(0) = 0

when I integrate, am I supposed to use the property below?

int du / (a^2 - u^2) = 1/2a ln(u+a / u-a) + c

or

how do I integrate this by using partial fraction?

tips anyone?


Homework Equations





The Attempt at a Solution

 
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You could use that integral to solve your integral. It would require some fancy footwork, so to speak, that is, getting your value of a correct.
 
You can integrate by partial fractions if you factor 9-4*x^2. Or you can use your formula after you do the u-substitution u=2*x. Your choice.
 
Dick said:
You can integrate by partial fractions if you factor 9-4*x^2. Or you can use your formula after you do the u-substitution u=2*x. Your choice.

using the property above, I get (1/12)ln((2x+3)(2x-3)) + c = t

then, e^12t = (2x+3)(2x-3) + e^c

e^12t - e^c = (2x+3)(2x-3)

here how can I go further to have x equals to whatever.

thanks a lot. it really helps.
 
shseo0315 said:
using the property above, I get (1/12)ln((2x+3)(2x-3)) + c = t

then, e^12t = (2x+3)(2x-3) + e^c

e^12t - e^c = (2x+3)(2x-3)

here how can I go further to have x equals to whatever.

thanks a lot. it really helps.

You've got two mistakes there. i) Shouldn't it be (2x+3)/(2x-3)? Not the product? And worse, ii) When you exponentiate e^(A+c) you get (e^A)*(e^c), not e^A+e^c. Do you see where you did that?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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