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I need to integrate this function

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

I´ve tried with partial integral with complex roots, but seems very complicated by this way.

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  • $\begingroup$ So this could be written as $$\frac {x-1}{x((x-1)^2+1)^2}$$ correct? It seems that using a $u$ substitution like $u=(x-1)^2+1$ might be interesting... $\endgroup$ – abiessu Jul 6 '14 at 0:32
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Hint:
Here's a start: $$\eqalign{\dfrac{x-1}{x(x^2-2x+2)^2}&=\dfrac{x}{x(x^2-2x+2)^2}-\dfrac1{x(x^2-2x+2)^2}\\&=\dfrac{1}{(x^2-2x+2)^2}-\dfrac1{x(x^2-2x+2)^2}.}$$ For the first fraction, a trigonometric substitution will do it and for the second fraction you still need some work on expanding it. Can you take it from here?

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Hint

If you use partial fraction decomposition, you should arrive to $$\dfrac{x-1}{x(x^2-2x+2)^2}=-\frac{1}{4 x}+\frac{x-2}{4 \left(x^2-2 x+2\right)}+\frac{x}{2 \left(x^2-2 x+2\right)^2}$$ The first term does not present any problem to integration; the third term is simple if you recognize something looking like $\frac{u'(x)}{u^2(x)}$. Concerning the second term, you can rewrite $$\frac{x-2}{x^2-2 x+2}=\frac{x-1}{x^2-2 x+2}-\frac{1}{x^2-2 x+2}$$

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As $\displaystyle\frac{d(x^2-2x+2)}{dx}=2(x-1),$

$\displaystyle\int\frac{x-1}{(x^2-2x+2)^2}\ dx=\int\frac{d(x^2-2x+2)}{2(x^2-2x+2)^2}=-\frac1{2(x^2-2x+2)}$

$$\implies\int\frac{x-1}{x(x^2-2x+2)^2}\ dx$$ $$=\frac1x\int\frac{x-1}{(x^2-2x+2)^2}\ dx-\int\left(\frac{d\dfrac1x}{dx}\int\frac{x-1}{(x^2-2x+2)^2}\ dx\right)dx$$

$$=-\frac1x\cdot\frac1{2(x^2-2x+2)}-\frac12\int\frac1{x^2(x^2-2x+2)}dx$$

Now using Partial Fraction Decomposition,

$\displaystyle\frac1{x^2(x^2-2x+2)}=\frac Ax+\frac B{x^2}+\frac{Cx+D}{x^2-2x+2}$

For, $\displaystyle\frac{Cx+D}{x^2-2x+2}=C\frac{(x-1)}{(x-1)^2+1^2}+(C+D)\frac1{(x-1)^2+1^2}$ where the last integral invites Trigonometric substitution

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