- Thread starter ol98
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\(\displaystyle \dfrac{du}{u(1-u)(a+u)} = dt\). Now integrate both sides. On the lhs I would use partial fraction. Post back showing your work.

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

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

Did you check it that way?

Where did

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You write: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

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

ln(u)-ln(1-u^2)/2 = t...........................where did '

2ln(u) - ln(1-u^2) = 2t

can you not cancel out the c's as they're on both sides?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

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No - those need not be equal.can

you not cancel out the c's on both side?

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=No - those need not be equal.

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ln(u)-ln(1-u^2)/2 + cYou 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

Let

c

Then:

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

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

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

Continue....