How do you understand the operations of equations and their solutions?

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hahaha uhhhh operations of e

Homework Statement


i'm reading this online math help thing, and i can't seem to memorize how/why this is

Homework Equations


eq0017L.gif

how did they get this?edit* how did they get the answer from the left to the right side of the arrow?

The Attempt at a Solution


i know this is prob. the beginner stuff, but i forgot this from a long time ago. and don't worry, this is only one step of the equation, not the whole answer.

Thanks!
 

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Is there a specific step(s) that you don't understand? All they are really doing is integrating both sides of the equation with respect to x and then isolating v(x) after.
 


They multiplied both sides by x^4
 


gabbagabbahey said:
Is there a specific step(s) that you don't understand? All they are really doing is integrating both sides of the equation with respect to x and then isolating v(x) after.

yeah, i got the integrating part, but how come in the next step, it went from x^-4 to x^4?


thanks!
 


I just told you, they multiplied both sides by x^4 ... and x^(-4) * x^4 = 1
 


NoMoreExams said:
I just told you, they multiplied both sides by x^4 ... and x^(-4) * x^4 = 1

oh haha yeah, thanks!


I was talking to "gabbagabbahey" in my last response though..
 
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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