1. ## integrate rational functions

Hi, im trying to integrate this function: (3x^2-10)/(x^2-4x+4)dx. I got it down to int 3+[(12x-22)/((x-2)^2 by long division. Then i get [A/(x-2)]+[B/(x-2)^2], i get stuck on trying to figure out what A and B are equal to. Can somone please help me with this?
thank you

2. Originally Posted by cowboys111
Hi, im trying to integrate this function: (3x^2-10)/(x^2-4x+4)dx. I got it down to int 3+[(12x-22)/((x-2)^2 by long division. Then i get [A/(x-2)]+[B/(x-2)^2], i get stuck on trying to figure out what A and B are equal to. Can somone please help me with this?
thank you
Get everything over the common denominator of (x - 2)^2 and equate the numerators of each side:

12x - 22 = A(x - 2) + B

=> 12x - 22 = Ax + (B - 2A)

=> A = 12 and -22 + B - 2A .......

3. We want: $
\int {\frac{{3 \cdot x^2 - 10}}
{{x^2 - 4x + 4}}dx}
$

Note that $
3x^2 - 10 = 3 \cdot \left( {x^2 - 4} \right) + 2 = 3 \cdot \left( {x - 2} \right) \cdot \left( {x + 2} \right) + 2
$
and $
x^2 - 4x + 4 = \left( {x - 2} \right)^2
$

Thus: $
\int {\frac{{3x^2 - 10}}
{{\left( {x - 2} \right)^2 }}} dx = \int {\frac{{3 \cdot \left( {x - 2} \right) \cdot \left( {x + 2} \right) + 2}}
{{\left( {x - 2} \right)^2 }}} dx = 3 \cdot \int {\frac{{x + 2}}
{{x - 2}}} dx + 2 \cdot \int {\frac{{dx}}
{{\left( {x - 2} \right)^2 }}} dx
$