# Thread: quick integration question...

1. ## quick integration question...

how does

$

\int \frac{-2x+1}{{x^2}+x+1}dx

$

become

$\int \frac{-2x-1}{{x^2}+x+1}dx\
+ 2 \int \frac{dx}{(x+{(1/2)^{2})}+(3/4)}
$

i got the completing the squares part but how does it turn to -1 and where did the 2 infront of the second integral come from...

2. $\int {\frac{{1 - 2x}}
{{x^2 + x + 1}}\,dx} = \int {\frac{{2 - \left( {x^2 + x + 1} \right)'}}
{{x^2 + x + 1}}\,dx} .$

Now splitting the original integral into two pieces, the second one is ready. (Since there's a hidden logarithm there, you may remove the absolute value bars for $x^2+x+1,$ 'cause it's always positive.) As for the first piece:

$\int {\frac{2}
{{x^2 + x + 1}}\,dx} = \int {\frac{8}
{{4x^2 + 4x + 4}}\,dx} = \int {\frac{8}
{{(2x + 1)^2 + 3}}\,dx} ,$

and this is an arctangent, let $u=2x+1$ and proceed from there.