Suppose `A=int (dx)/(x^(2)+6x+25)` and `B=int(dx)/(x^(2)-6x-27)`.
If `12(A+B)=lambda*tan^(-1)((x+3)/(4))+mu* ln |(x-9)/(x+3)|+C`, then the value of `(lambda+mu)` is ... .

Answer

To solve the problem, we need to evaluate the integrals \( A \) and \( B \) and then find the values of \( \lambda \) and \( \mu \) from the expression \( 12(A+B) \).

### Step 1: Evaluate \( A = \int \frac{dx}{x^2 + 6x + 25} \)

1. **Complete the square** for the quadratic in the denominator:
\[ x^2 + 6x + 25 = (x^2 + 6x + 9) + 16 = (x + 3)^2 + 4^2 \]
Thus, we rewrite \( A \):
\[ A = \int \frac{dx}{(x + 3)^2 + 4^2} \]

2. **Use the formula** for the integral of the form \( \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \):
Here, \( a = 4 \) and \( x \) is replaced by \( x + 3 \):
\[ A = \frac{1}{4} \tan^{-1} \left( \frac{x + 3}{4} \right) + C \]

### Step 2: Evaluate \( B = \int \frac{dx}{x^2 - 6x - 27} \)

1. **Complete the square** for the quadratic in the denominator:
\[ x^2 - 6x - 27 = (x^2 - 6x + 9) - 36 = (x - 3)^2 - 36 \]
Thus, we rewrite \( B \):
\[ B = \int \frac{dx}{(x - 3)^2 - 6^2} \]

2. **Use the formula** for the integral of the form \( \int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \ln \left| \frac{x - a}{x + a} \right| + C \):
Here, \( a = 6 \) and \( x \) is replaced by \( x - 3 \):
\[ B = \frac{1}{12} \ln \left| \frac{(x - 3) - 6}{(x - 3) + 6} \right| + C = \frac{1}{12} \ln \left| \frac{x - 9}{x + 3} \right| + C \]

### Step 3: Combine \( A \) and \( B \)

Now we have:
\[ A = \frac{1}{4} \tan^{-1} \left( \frac{x + 3}{4} \right) + C_1 \]
\[ B = \frac{1}{12} \ln \left| \frac{x - 9}{x + 3} \right| + C_2 \]

Thus,
\[ A + B = \frac{1}{4} \tan^{-1} \left( \frac{x + 3}{4} \right) + \frac{1}{12} \ln \left| \frac{x - 9}{x + 3} \right| + C \]

### Step 4: Calculate \( 12(A + B) \)

\[ 12(A + B) = 12 \left( \frac{1}{4} \tan^{-1} \left( \frac{x + 3}{4} \right) + \frac{1}{12} \ln \left| \frac{x - 9}{x + 3} \right| \right) \]
\[ = 3 \tan^{-1} \left( \frac{x + 3}{4} \right) + \ln \left| \frac{x - 9}{x + 3} \right| + C \]

### Step 5: Compare with the given expression

The given expression is:
\[ 12(A + B) = \lambda \tan^{-1} \left( \frac{x + 3}{4} \right) + \mu \ln \left| \frac{x - 9}{x + 3} \right| + C \]

From this, we can identify:
- \( \lambda = 3 \)
- \( \mu = 1 \)

### Step 6: Find \( \lambda + \mu \)

\[ \lambda + \mu = 3 + 1 = 4 \]

### Final Answer
The value of \( \lambda + \mu \) is \( 4 \).