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I'am trying to integrate the following function$\int \frac{2}{(v-1)^2}dv$ So far I've tried $u$ substitution.Here are the steps.Let $u = (v-1)^2$ then$\frac{du}{dv} = 2v-2$ solving for $dv$ $dv = \frac{1}{2(v-1)}du$ This doesn't seem to me a sensible way of working out this integration.I'am stuck Please help.

3 Answers 3

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Let $u=v-1$, $du=dv$. Then

$\int\frac{dv}{(v-1)^2}=\int\frac{du}{u^2}.$

I'm sure you can finish, now!

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    I do my best! :-)2011-12-27
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Letting $u=v-1$ makes the integral $\int2u^{-2}du$, which I bet you can do on your own.

Always consider the possibility of the simplest substitutions (e.g. linear) before anything else.

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Are you familiar with the power rule for integration, $\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\quad (n\neq 1)\quad ?$ There is a different $u$-substitution you can make where you can apply it.

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    yes i'am familiar2011-12-27