Why do we choose to multiply by 2 in simplifying this equation?

[trajan22]

the main problem I am having is i can't figure out how they simplified this equation. Ill try and use latex but its my first time so i don't know if it will work.
<br /> \int(1-\frac{1}{3(x+1)}+\frac{(x-2)}{3(x^2-x+1)}) dx<br />
and somehow this simplifies to this
<br /> x- (1/3)ln(abs(x+1)) +(1/6) \int \frac{(2x-1)}{(x^2-x+1)} dx - (1/2) \int \frac{(dx)}{(x-.5)^2+(3/4)} <br />
i understand how to get the x-(1/3)ln(2x-1) but i don't get how they separated the other two terms for integration. or how they pulled out (1/6)

 

[neutrino]

Given <br /> \int\frac{(x-2)}{3(x^2-x+1)}\, dx<br />, multiply the numerator and denominator by 2 and rewrite the numerator as (2x-1)-3.

 

[cristo]

Ok, well the third term in the integrand of the original question is <br /> \frac{(x-2)}{3(x^2-x+1)}=\frac{(2x-4)}{6(x^2-x+1)}=\frac{(2x-1)}{6(x^2-x+1)}-\frac{3}{6(x^2-x+1)}=\frac{1}{6}\frac{(2x-1)}{(x^2-x+1)}-\frac{1}{2}\frac{1}{(x^2-x+1)}=\frac{1}{6}\frac{(2x-1)}{(x^2-x+1)}-\frac{1}{2}\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}} where the last step in the denominator of the second term is done by completing the square.

 

[trajan22]

ok i understand what is going on now, but why do we chose to multiply the numerator and denominator by 2, and why do we chose 2 rather than any other number.

 

[GeoMike]

ok i understand what is going on now, but why do we chose to multiply the numerator and denominator by 2, and why do we chose 2 rather than any other number.

Multiplying by 2 allows the fraction to be rewritten so that part of it is in the form du/u, which can be integrated to to get ln(abs(u))

-GeoMike-

 

[neutrino]

ok i understand what is going on now, but why do we chose to multiply the numerator and denominator by 2, and why do we chose 2 rather than any other number.

Why don't you tell us what you arrive at after using some other number/method? If it's more effective, we'll ditch the earlier one.

At an glance, you might think that this is solvable by the substitution u = x²- x +1, but once you do it, you'll realize where the 2 comes from.