int 3/((x-2)(x+1))dx∫3(x−2)(x+1)dx
The method of partial fractions involves splitting the denominator into its factors. Lets ignore the integral for now. We are looking for two values, AA and BB such that;
A/(x-2) + B/(x+1) = 3/((x-2)(x+1))Ax−2+Bx+1=3(x−2)(x+1)
Now we need to combine AA and BB over a common denominator.
A/((x-2))((x+1))/((x+1)) + B/((x+1)) ((x-2))/((x-2)) = 3/((x-2)(x+1))A(x−2)(x+1)(x+1)+B(x+1)(x−2)(x−2)=3(x−2)(x+1)
(A(x+1)+B(x-2))/((x-2)(x+1)) = 3/((x-2)(x+1))A(x+1)+B(x−2)(x−2)(x+1)=3(x−2)(x+1)
We can cancel the denominator on both sides, leaving;
A(x+1)+B(x-2) = 3A(x+1)+B(x−2)=3
Now we can solve for AA and BB by selectively choosing our values of xx so that the terms in parenthesis are 00. Let x=-1x=−1.
Acolor(red)cancel(color(black)((-1+1)))^0 + B(-1-2) = 3
-3B = 3
B=-1
Let x=2.
A(2+1)+Bcolor(red)cancel(color(black)((2-2)))^0 = 3
3A = 3
A=1
Now that we have our values for A and B we can split apart our integral by plugging them in.
int 3/((x-2)(x+1)) dx = int (1/(x-2) +(- 1)/(x+1)) dx
=int 1/(x-2) dx - int 1/(x+1) dx
Now we have two integrals that we can solve using substitution. Let u_1=x-2 for the first integral and u_2=x+1 for the second.
int 1/u_1 du_1 - int 1/u_2 du_2
ln(u_1) - ln(u_2) +C
ln((u_1)/(u_2)) + C
Plug in the values for u_1 and u_2.
=ln((x-2)/(x+1)) +C
