Efficient Integration: v/(v^2 + 4) Simplified

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I know that if it were 1/(v2+4) then it would be ln(v2+4) but the other v in the numberator is throwing me off, any help would be much appreciated!
 
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Never mind, I figured it out!
 
Aaron Curran said:
I know that if it were 1/(v2+4) then it would be ln(v2+4)

No, it wouldn't. Can you see why?

but the other v in the numberator is throwing me off, any help would be much appreciated!

There is no such thing as a "numberator". The proper terms for the parts of a fraction are "numerator" and "denominator". You should have learned this long before starting calculus.
 
SteamKing said:
No, it wouldn't. Can you see why?
There is no such thing as a "numberator". The proper terms for the parts of a fraction are "numerator" and "denominator". You should have learned this long before starting calculus.

It was a typo haha, I've worked it out now anyway, feel very stupid for making this thread :sorry:
 
Aaron Curran said:
I know that if it were 1/(v2+4) then it would be ln(v2+4)
No, that is incorrect. You can verify what I said by differentiating ln(v2 + 4), which does not result in 1/(v2 + 4).
Aaron Curran said:
but the other v in the numberator is throwing me off, any help would be much appreciated!
 
Don't tell us you have got the answer, tell us the answer you have got!
 
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Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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