Integration Proof

ol98

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Anyone able to prove this integral?
 

Jomo

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Yes, many people on this forum can solve this integral. The important question is can you solve it? What have you tried? Where are you stuck?

\(\displaystyle \dfrac{du}{u(1-u)(a+u)} = dt\). Now integrate both sides. On the lhs I would use partial fraction. Post back showing your work.
 

ol98

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log(u) - 1/2 log(1-u^2)=t ?
 

Subhotosh Khan

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log(u) - 1/2 log(1-u^2)=t ?
Why do you have a "?" with your answer?

Integration is easy to check - by differentiating the answer and comparing with the original function.

Did you check it that way?

Where did the constant of integration go?
 

ol98

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Not entirely sure if I'm on the right path, so my working so far:

ln(u)-ln(1-u^2)/2 + c = t + c
ln(u)-ln(1-u^2)/2 = t
2ln(u) - ln(1-u^2) = 2t
ln(u^2/(1-u^2) = 2t
u^2/(1-u^2) = e^2t

Any feedback on that working would be really helpful, thanks
 

Subhotosh Khan

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Not entirely sure if I'm on the right path, so my working so far:

ln(u)-ln(1-u^2)/2 + c = t + c
ln(u)-ln(1-u^2)/2 = t
2ln(u) - ln(1-u^2) = 2t
ln(u^2/(1-u^2) = 2t
u^2/(1-u^2) = e^2t

Any feedback on that working would be really helpful, thanks
You write:

ln(u)-ln(1-u^2)/2 + c = t + c
ln(u)-ln(1-u^2)/2 = t...........................where did 'c' go?
2ln(u) - ln(1-u^2) = 2t
 

ol98

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You write:

ln(u)-ln(1-u^2)/2 + c = t + c
ln(u)-ln(1-u^2)/2 = t...........................where did 'c' go?
2ln(u) - ln(1-u^2) = 2t
can you not cancel out the c's as they're on both sides?
 

Subhotosh Khan

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ol98

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No - those need not be equal.
you ok to explain what you mean by this, because how do I look to carry on the proof if I'm trying to get terms on one side so I have u=
 

Subhotosh Khan

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You write:

ln(u)-ln(1-u^2)/2 + c = t + c
ln(u)-ln(1-u^2)/2 = t...........................where did 'c' go?
2ln(u) - ln(1-u^2) = 2t
ln(u)-ln(1-u^2)/2 + c1 = t + c2

Let

c1 - c2 = c3 = ln(c)

Then:

ln(u)-ln(1-u^2)/2 + c1 = t + c2

ln(u)-ln(1-u^2)/2 + ln(c) = t

c*u/(1-u^2) = e^t

Continue....
 
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