Integration Proof

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?

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