# Techniques of integration (4)

• May 17th 2009, 05:35 PM
NonCommAlg
Techniques of integration (4)
Evaluate $\displaystyle I=\int \frac{x(2x-1)}{(x^2-x+1)^2 + 1} \ dx.$
• May 18th 2009, 03:14 AM
simplependulum
I think you wanna test whether we are brave enough to expand and factorize the denominator (Nod)

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

and

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

$\displaystyle = \frac{-x}{x^2 + 1} + \frac{x}{x^2 - 2x + 2}$
• May 18th 2009, 03:58 AM
NonCommAlg
Quote:

Originally Posted by simplependulum
I think you wanna test whether we are brave enough to expand and factorize the denominator (Nod)

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

and

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

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

and you were brave enough! well done!