Question

itegrate 3/(x+3)(x-2) dx

Answer 1

To integrate 3/(x+3)(x-2) dx, we need to use partial fraction decomposition.

First, we need to rewrite the expression as a sum of two fractions:

3/(x+3)(x-2) = A/(x+3) + B/(x-2)

Multiplying both sides by (x+3)(x-2), we get:
3 = A(x-2) + B(x+3)

Expanding the right side gives:
3 = Ax - 2A + Bx + 3B
3 = x(A + B) + 3B - 2A

Now, we can equate coefficients of like terms to get the values of A and B:

A + B = 0
-2A + 3B = 3

From the first equation, we get:
A = -B

Substituting this into the second equation:
-2(-B) + 3B = 3
2B + 3B = 3
5B = 3
B = 3/5

Now we can find A:
A = -B = -3/5

Therefore, the partial fraction decomposition is:
3/(x+3)(x-2) = -3/(5(x+3)) + 3/5(x-2)

Integrating the decomposed fractions, we get:
∫ -3/(5(x+3)) dx + ∫ 3/(5(x-2)) dx
= -3/5 * ln|x+3|/5 + 3/5 * ln|x-2| + C

where C is the constant of integration.