Evaluate :
`int (dx)/((2x+1)(x^(2)+4))`

Answer

To evaluate the integral

\[ I = \int \frac{dx}{(2x+1)(x^2+4)}, \]

we will use the method of partial fractions.

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

We can express the integrand as:

\[ \frac{1}{(2x+1)(x^2+4)} = \frac{A}{2x+1} + \frac{Bx+C}{x^2+4}, \]

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

### Step 2: Combine the fractions

Multiplying both sides by the denominator \((2x+1)(x^2+4)\) gives:

\[ 1 = A(x^2 + 4) + (Bx + C)(2x + 1). \]

Expanding the right-hand side:

\[ 1 = Ax^2 + 4A + (2Bx^2 + Bx + 2Cx + C). \]

Combining like terms, we have:

\[ 1 = (A + 2B)x^2 + (B + 2C)x + (4A + C). \]

### Step 3: Set up a system of equations

Now, we can equate the coefficients from both sides:

1. For \(x^2\): \(A + 2B = 0\)
2. For \(x\): \(B + 2C = 0\)
3. Constant term: \(4A + C = 1\)

### Step 4: Solve the system of equations

From the first equation, we can express \(A\) in terms of \(B\):

\[ A = -2B. \]

Substituting \(A\) into the third equation:

\[ 4(-2B) + C = 1 \implies -8B + C = 1 \implies C = 1 + 8B. \]

Now substituting \(C\) into the second equation:

\[ B + 2(1 + 8B) = 0 \implies B + 2 + 16B = 0 \implies 17B + 2 = 0 \implies B = -\frac{2}{17}. \]

Now substituting \(B\) back to find \(A\) and \(C\):

\[ A = -2\left(-\frac{2}{17}\right) = \frac{4}{17}, \]
\[ C = 1 + 8\left(-\frac{2}{17}\right) = 1 - \frac{16}{17} = \frac{1}{17}. \]

### Step 5: Write the partial fraction decomposition

Now we can write:

\[ \frac{1}{(2x+1)(x^2+4)} = \frac{4/17}{2x+1} + \frac{-2/17 x + 1/17}{x^2+4}. \]

### Step 6: Integrate each term

Now we can integrate term by term:

\[ I = \int \left(\frac{4/17}{2x+1} + \frac{-2/17 x + 1/17}{x^2+4}\right) dx. \]

This gives us:

\[ I = \frac{4}{17} \int \frac{1}{2x+1} dx - \frac{2}{17} \int \frac{x}{x^2+4} dx + \frac{1}{17} \int \frac{1}{x^2+4} dx. \]

### Step 7: Solve each integral

1. The first integral:

\[ \int \frac{1}{2x+1} dx = \frac{1}{2} \ln |2x+1| + C. \]

2. The second integral can be solved using substitution \(u = x^2 + 4\):

\[ \int \frac{x}{x^2+4} dx = \frac{1}{2} \ln |x^2 + 4| + C. \]

3. The third integral:

\[ \int \frac{1}{x^2+4} dx = \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C. \]

### Step 8: Combine the results

Putting it all together, we have:

\[ I = \frac{4}{17} \cdot \frac{1}{2} \ln |2x+1| - \frac{2}{17} \cdot \frac{1}{2} \ln |x^2 + 4| + \frac{1}{17} \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C. \]

### Final Result

Thus, the evaluated integral is:

\[ I = \frac{2}{17} \ln |2x+1| - \frac{1}{17} \ln |x^2 + 4| + \frac{1}{34} \tan^{-1}\left(\frac{x}{2}\right) + C. \]