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How do you integrate $$ \int {1\over x^2(x-1)^3} \, dx $$

I'd love some general advice on how to approach problems like these. I tried partial fraction expansion, but in the end it got me nowhere.

Let me clarify: This is where partial fraction expansion got me:

$$ \int {3\over (x-1)} \, dx - \int {2\over(x-1)^2} \, dx + \int {2\over(x-1)^3} \, dx - \int {3x-1\over x^2} \, dx $$

Problem is I still don't know how to integrate second and third fraction. And I feel I probably messed up somewhere along the road.

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    $\begingroup$ Hint: Make the substitution $u =x - 1$ for the last integrals when needed. $\endgroup$ – Ali Caglayan Jan 1 '15 at 22:20
  • $\begingroup$ @Alizter perfect, I totally forgot about u-sub! $\endgroup$ – user1904218 Jan 1 '15 at 22:27
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Using partial fractions, you'll find

$$\frac{1}{x^2(x-1)^3} = -\frac{1}{x^2} - \frac{3}{x} + \frac{3}{x-1} - \frac{2}{(x-1)^2} + \frac{1}{(x-1)^3}.$$

EDIT: To integrate the fractions in your original post you claim to have trouble with, remember that

$$\left(\frac{1}{x} \right)' = - \frac{1}{x^2} \text{ and } \left(\frac{1}{x^2} \right)' = - \frac{2}{x^3}.$$

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  • $\begingroup$ I'm sorry, that was a typo. I meant the second and third! But thank you very much, seems like I need to use u-sub for the seoncd and third fractions. $\endgroup$ – user1904218 Jan 1 '15 at 22:26
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    $\begingroup$ @user1904218: You don't actually (explicitly) need substitution, you can use the chain rule which gives $((x-1)^{-1})' = -(x-1)^{-2}$ immediately. $\endgroup$ – Huy Jan 1 '15 at 22:27
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hint: use that $$\frac{1}{x^2(x-1)^3}=3\, \left( x-1 \right) ^{-1}-{x}^{-2}+ \left( x-1 \right) ^{-3}-2\, \left( x-1 \right) ^{-2}-3\,{x}^{-1} $$

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