Using a fraction whose denominator is quadratic, you need to check whether or not it is factorable.
6x^2 + 5x - 6 = (2x+3)(3x-2)6x2+5x−6=(2x+3)(3x−2)
Since it is, we only need to write out the following:
int (13x)/((2x + 3)(3x - 2))dx = int A/(2x + 3) + B/(3x - 2)dx∫13x(2x+3)(3x−2)dx=∫A2x+3+B3x−2dx
To find what AA and BB are, you could start by cross-multiplying. For now, ignore the integral sign and dxdx.
= (A(3x-2) + B(2x+3))/((2x+3)(3x - 2))=A(3x−2)+B(2x+3)(2x+3)(3x−2)
= (3Ax - 2A + 2Bx+3B)/((2x+3)(3x - 2))=3Ax−2A+2Bx+3B(2x+3)(3x−2)
= ((3A + 2B)x + (-2A + 3B))/((2x+3)(3x - 2))=(3A+2B)x+(−2A+3B)(2x+3)(3x−2)
Thus, we have the equations:
-2A + 3B = 0 => B = 2/3 A−2A+3B=0⇒B=23A
3A + 2B = 13 => 3A + 4/3 A = 133A+2B=13⇒3A+43A=13
13/3 A = 13133A=13
:. color(green)(A = 3) => color(green)(B = 2)
At this point we are pretty much done.
= int 3/(2x + 3) + 2/(3x - 2)dx
= color(blue)(3/2ln|2x + 3| + 2/3ln|3x - 2| + C)
