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

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.

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

$$\frac{1}{x^2(x-1)^3} = -\frac{1}{x^2} - \frac{3}{x} + \frac{3}{x-1} - \frac{2}{(x-1)^2} + \frac{1}{(x-1)^3}.$$
$$\left(\frac{1}{x} \right)' = - \frac{1}{x^2} \text{ and } \left(\frac{1}{x^2} \right)' = - \frac{2}{x^3}.$$
• @user1904218: You don't actually (explicitly) need substitution, you can use the chain rule which gives $((x-1)^{-1})' = -(x-1)^{-2}$ immediately. – Huy Jan 1 '15 at 22:27
hint: use that $$\frac{1}{x^2(x-1)^3}=3\, \left( x-1 \right) ^{-1}-{x}^{-2}+ \left( x-1 \right) ^{-3}-2\, \left( x-1 \right) ^{-2}-3\,{x}^{-1}$$