# how to integrate$\frac{(x-1)}{ x(x^2-2x+2)^2}$ ??

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.

• 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... – abiessu Jul 6 '14 at 0:32

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?

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 $$\frac{x-2}{x^2-2 x+2}=\frac{x-1}{x^2-2 x+2}-\frac{1}{x^2-2 x+2}$$

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