Integration

**Stuck Man** · January 5th 2012, 06:51 AM

Is this question answered ok or should I have an equation in t?

**Siron** · January 5th 2012, 07:25 AM

The problem is the fact that you write:
$\left(\frac{v}{2}-\frac{v^2}{2}\right)^{-1}=\frac{2}{v}-\frac{2}{v^2}$
which is incorrect.

You can say that:
$\left(\frac{v}{2}-\frac{v^2}{2}\right)^{-1}=\left(\frac{v-v^2}{2}\right)^{-1}=\frac{2}{v-v^2}$

**Stuck Man** · January 5th 2012, 08:21 AM

Is this correct now? I noticed there is another way of doing the integral but I have not learnt about arctanh.

**Siron** · January 5th 2012, 08:28 AM

It looks good now!

You can also apply 'complete the square' to solve $\int \frac{dv}{v^2-v}$

**Stuck Man** · January 5th 2012, 08:36 AM

Thats what I found. I have made an equation in t from the final equation. v=(1-4e^(t/2))^-1

**Siron** · January 5th 2012, 08:45 AM

It's correct!

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