Evaluate: ∫(2x + 9)/(x + 2)(x - 3)^2dx ​

Question

Evaluate:

\(\int\frac{(2x+9)}{(x+2)(x-3)^2}dx\)

∫(2x + 9)/(x + 2)(x - 3)²dx

Answer 1

Let I = \(\int\frac{(2x+9)}{(x+2)(x-3)^2}dx\)

Now putting, \(\frac{2x+9}{(x+2)(x-3)^2}\) = \(\frac{A}{(x+2)}\) + \(\frac{B}{(x-3)}\) + \(\frac{C}{(x-3)^2}\)......(1)

A(x - 3)² + B(x + 2)(x - 3) + C(x + 2) = 2x + 9 

Now put x - 3 = 0 

Therefore, x = 3 

A(0) + B(0) + C(3 + 2) = 6 + 9 =15 

C = 3 

Now put x + 2 = 0 

Therefore, x = -2 

A(- 2 - 3)² + B(0) + C(0) = - 4 + 9 = 5

A = 1/5

Equating the coefficient of x²,we get,

A + B = 0

1/5 + B = 0

B = - 1/5

From equation (1), we get,