`int (1)/((x+b)(x^(2)+a^(2)))dx`

Answer

To solve the integral \( \int \frac{1}{(x+b)(x^2+a^2)} \, dx \), we will use the method of partial fractions. Here’s the step-by-step solution:

### Step 1: Set up the partial fraction decomposition
We can express the integrand as:
\[ \frac{1}{(x+b)(x^2+a^2)} = \frac{A}{x+b} + \frac{Bx+C}{x^2+a^2} \]
where \( A \), \( B \), and \( C \) are constants to be determined.

### Step 2: Clear the denominators
Multiply both sides by the denominator \( (x+b)(x^2+a^2) \):
\[ 1 = A(x^2 + a^2) + (Bx + C)(x + b) \]

### Step 3: Expand the right-hand side
Expanding the right-hand side gives:
\[ 1 = A(x^2 + a^2) + Bx^2 + Bbx + Cx + Cb \]
Combining like terms, we have:
\[ 1 = (A + B)x^2 + (Bb + C)x + (Aa^2 + Cb) \]

### Step 4: Set up the system of equations
Now, we equate the coefficients from both sides:
1. Coefficient of \( x^2 \): \( A + B = 0 \)
2. Coefficient of \( x \): \( Bb + C = 0 \)
3. Constant term: \( Aa^2 + Cb = 1 \)

### Step 5: Solve the system of equations
From the first equation, we can express \( B \) in terms of \( A \):
\[ B = -A \]

Substituting \( B = -A \) into the second equation:
\[ -bA + C = 0 \implies C = bA \]

Substituting \( B \) and \( C \) into the third equation:
\[ Aa^2 + bA \cdot b = 1 \implies A(a^2 + b^2) = 1 \implies A = \frac{1}{a^2 + b^2} \]

Now substituting back to find \( B \) and \( C \):
\[ B = -\frac{1}{a^2 + b^2}, \quad C = \frac{b}{a^2 + b^2} \]

### Step 6: Rewrite the integral
Now we can rewrite the integral:
\[ \int \left( \frac{1}{a^2 + b^2} \cdot \frac{1}{x+b} - \frac{1}{a^2 + b^2} \cdot \frac{x}{x^2 + a^2} + \frac{b}{a^2 + b^2} \cdot \frac{1}{x^2 + a^2} \right) dx \]

### Step 7: Integrate each term
1. The first term:
\[ \int \frac{1}{a^2 + b^2} \cdot \frac{1}{x+b} \, dx = \frac{1}{a^2 + b^2} \ln |x+b| \]

2. The second term:
\[ -\frac{1}{a^2 + b^2} \int \frac{x}{x^2 + a^2} \, dx = -\frac{1}{2(a^2 + b^2)} \ln |x^2 + a^2| \]

3. The third term:
\[ \frac{b}{a^2 + b^2} \int \frac{1}{x^2 + a^2} \, dx = \frac{b}{a^2 + b^2} \cdot \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \]

### Step 8: Combine the results
Putting it all together, we have:
\[ \int \frac{1}{(x+b)(x^2+a^2)} \, dx = \frac{1}{a^2 + b^2} \ln |x+b| - \frac{1}{2(a^2 + b^2)} \ln |x^2 + a^2| + \frac{b}{a(a^2 + b^2)} \tan^{-1} \left( \frac{x}{a} \right) + C \]

### Final Answer
Thus, the final result is:
\[ \int \frac{1}{(x+b)(x^2+a^2)} \, dx = \frac{1}{a^2 + b^2} \ln |x+b| - \frac{1}{2(a^2 + b^2)} \ln |x^2 + a^2| + \frac{b}{a(a^2 + b^2)} \tan^{-1} \left( \frac{x}{a} \right) + C \]