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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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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