Techniques of integration (4)

**NonCommAlg** · May 17th 2009, 05:35 PM

Evaluate $I=\int \frac{x(2x-1)}{(x^2-x+1)^2 + 1} \ dx.$

**simplependulum** · May 18th 2009, 03:14 AM

I think you wanna test whether we are brave enough to expand and factorize the denominator

$(x^2 - x + 1)^2 + 1$
$= x^4 -2x^3 + 3x^2 - 2x + 1 + 1$
$= x^4 -2x^3 + 3x^2 - 2x + 2$
$= ( x^4 + 3x^2 + 2 ) - 2x(x^2 + 1)$
$= ( x^2 + 1 )(x^2 + 2) - 2x( x^2 + 1)$
$= (x^2 + 1)(x^2 - 2x + 2)$

and

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

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

**NonCommAlg** · May 18th 2009, 03:58 AM

Originally Posted by simplependulum

I think you wanna test whether we are brave enough to expand and factorize the denominator

$(x^2 - x + 1)^2 + 1$
$= x^4 -2x^3 + 3x^2 - 2x + 1 + 1$
$= x^4 -2x^3 + 3x^2 - 2x + 2$
$= ( x^4 + 3x^2 + 2 ) - 2x(x^2 + 1)$
$= ( x^2 + 1 )(x^2 + 2) - 2x( x^2 + 1)$
$= (x^2 + 1)(x^2 - 2x + 2)$

and

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

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

and you were brave enough! well done!

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