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Can we solve this integral using a good change of variable?

The integral is:

$$\int\frac{1}{x^2(x-1)^3}\mathrm{d}x$$

I know how to solve using the way of decomposition, but is there another way?

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    $\begingroup$ $z=\frac{1}{x}-1$ $\endgroup$ – Hari Shankar Oct 28 '17 at 7:48
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Your integral is of general forms called Differential Binomial. There are just 3 possible cases in which you can express the integral as elementary functions (see this). According to it, what the other answer suggested is applicable. It means that you can think on $$t=-1+x.$$ In fact, $a=-1,b=1,p=-3,m=-2$ while $(m+1)/n=-1\in\mathbb{Z}.$

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use that $$\frac{1}{x^2(x-1)^3}=\left( x-1 \right) ^{-3}-2\, \left( x-1 \right) ^{-2}-{x}^{-2}+3\, \left( x-1 \right) ^{-1}-3\,{x}^{-1} $$

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