Evaluate :
`int(dx)/((2-x)(x^(2)+3))`

Answer

To evaluate the integral

\[ \int \frac{dx}{(2-x)(x^2+3)}, \]

we will use the method of partial fractions. Let's go through the steps:

### Step 1: Set up the partial fraction decomposition

We want to express the integrand as a sum of simpler fractions. We can write:

\[ \frac{1}{(2-x)(x^2+3)} = \frac{A}{2-x} + \frac{Bx + C}{x^2 + 3}, \]

where \(A\), \(B\), and \(C\) are constants we need to determine.

### Step 2: Combine the right-hand side

Multiplying both sides by the common denominator \((2-x)(x^2+3)\) gives:

\[ 1 = A(x^2 + 3) + (Bx + C)(2 - x). \]

### Step 3: Expand the right-hand side

Expanding the right-hand side:

\[ 1 = A(x^2 + 3) + (2B - C)x + (2C - B)x^2. \]

Combining like terms, we have:

\[ 1 = (A - B)x^2 + (2B - C)x + (3A + 2C). \]

### Step 4: Set up the system of equations

Now we can equate the coefficients from both sides:

1. For \(x^2\): \(A - B = 0\)
2. For \(x\): \(2B - C = 0\)
3. For the constant term: \(3A + 2C = 1\)

### Step 5: Solve the system of equations

From the first equation, we have:

\[ A = B. \]

Substituting \(A\) for \(B\) in the second equation:

\[ 2A - C = 0 \implies C = 2A. \]

Now substituting \(B = A\) and \(C = 2A\) into the third equation:

\[ 3A + 2(2A) = 1 \implies 3A + 4A = 1 \implies 7A = 1 \implies A = \frac{1}{7}. \]

Thus,

\[ B = \frac{1}{7}, \quad C = 2A = \frac{2}{7}. \]

### Step 6: Write the partial fraction decomposition

Now we can write the integrand as:

\[ \frac{1}{(2-x)(x^2+3)} = \frac{1/7}{2-x} + \frac{(1/7)x + (2/7)}{x^2 + 3}. \]

### Step 7: Integrate each term

Now we can integrate each term separately:

\[ \int \left( \frac{1/7}{2-x} + \frac{(1/7)x + (2/7)}{x^2 + 3} \right) dx = \frac{1}{7} \int \frac{1}{2-x} dx + \frac{1}{7} \int \frac{x}{x^2 + 3} dx + \frac{2}{7} \int \frac{1}{x^2 + 3} dx. \]

### Step 8: Compute the integrals

1. The first integral:

\[ \frac{1}{7} \int \frac{1}{2-x} dx = -\frac{1}{7} \ln |2-x|. \]

2. The second integral requires substitution. Let \(u = x^2 + 3\), then \(du = 2x \, dx\) or \(\frac{du}{2} = x \, dx\):

\[ \frac{1}{7} \int \frac{x}{x^2 + 3} dx = \frac{1}{14} \ln |x^2 + 3|. \]

3. The third integral can be computed using the formula for the integral of \( \frac{1}{x^2 + a^2} \):

\[ \frac{2}{7} \int \frac{1}{x^2 + 3} dx = \frac{2}{7} \cdot \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right). \]

### Step 9: Combine the results

Putting it all together, we have:

\[ \int \frac{dx}{(2-x)(x^2+3)} = -\frac{1}{7} \ln |2-x| + \frac{1}{14} \ln |x^2 + 3| + \frac{2}{7\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + C. \]

### Final Answer

Thus, the final result is:

\[ \int \frac{dx}{(2-x)(x^2+3)} = -\frac{1}{7} \ln |2-x| + \frac{1}{14} \ln |x^2 + 3| + \frac{2}{7\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + C. \]

---