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I'm trying to use the "change of variable" technique to integrate this function.

I have this function

$I_1 = \int_0^{\infty} \frac{\alpha \mu x (1+\mu x)^{-\alpha -1}}{1-\lambda x (1+\mu x)^{-\alpha }}dx$

To integrate this, one way is to use the change of variable technique. I see that

$u = (1 +\mu x)^{-\alpha }$

and

$du = -\alpha \mu (\mu x+1)^{-\alpha -1} dx$

so now,

$I_2 = -\int \frac{x du}{1-\lambda x u}$.

But, there is still an "extra" $x$.

Is it possible to do "change of variables" twice?

How can this be done?

1 Answers 1

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Use the fact that $$x=\frac{1}{\mu}(u^{-\frac{1}{\alpha}}-1)$$

Hope this helps.