`I=int(5x+4)/(x^(2)+2x+5)dx.`

Answer

To solve the integral \( I = \int \frac{5x + 4}{x^2 + 2x + 5} \, dx \), we will follow these steps:

### Step 1: Simplify the Integral
We start with the integral:
\[ I = \int \frac{5x + 4}{x^2 + 2x + 5} \, dx \]
Here, we can notice that the denominator \( x^2 + 2x + 5 \) can be rewritten to facilitate integration.

### Step 2: Complete the Square
The quadratic expression in the denominator can be completed to a perfect square:
\[ x^2 + 2x + 5 = (x + 1)^2 + 4 \]
Thus, we rewrite the integral as:
\[ I = \int \frac{5x + 4}{(x + 1)^2 + 4} \, dx \]

### Step 3: Split the Integral
Next, we can split the integral into two parts:
\[ I = \int \frac{5x + 2 + 2}{(x + 1)^2 + 4} \, dx = \int \frac{5x + 2}{(x + 1)^2 + 4} \, dx + \int \frac{2}{(x + 1)^2 + 4} \, dx \]

### Step 4: Substitute for the First Integral
For the first integral, we can use the substitution \( u = x + 1 \), which gives \( du = dx \) and \( x = u - 1 \):
\[ I_1 = \int \frac{5(u - 1) + 2}{u^2 + 4} \, du = \int \frac{5u - 3}{u^2 + 4} \, du \]
This can be split into:
\[ I_1 = 5 \int \frac{u}{u^2 + 4} \, du - 3 \int \frac{1}{u^2 + 4} \, du \]

### Step 5: Solve the First Integral
For the first part \( \int \frac{u}{u^2 + 4} \, du \), we can use the substitution \( v = u^2 + 4 \) leading to:
\[ dv = 2u \, du \implies du = \frac{dv}{2u} \]
Thus,
\[ \int \frac{u}{u^2 + 4} \, du = \frac{1}{2} \ln |u^2 + 4| + C \]

### Step 6: Solve the Second Integral
For the second part \( \int \frac{1}{u^2 + 4} \, du \):
\[ \int \frac{1}{u^2 + 4} \, du = \frac{1}{2} \tan^{-1}\left(\frac{u}{2}\right) + C \]

### Step 7: Combine the Results
Combining these results gives:
\[ I_1 = 5 \left( \frac{1}{2} \ln |u^2 + 4| \right) - 3 \left( \frac{1}{2} \tan^{-1}\left(\frac{u}{2}\right) \right) \]
Substituting back \( u = x + 1 \):
\[ I_1 = \frac{5}{2} \ln |(x + 1)^2 + 4| - \frac{3}{2} \tan^{-1}\left(\frac{x + 1}{2}\right) \]

### Step 8: Solve the Second Integral
Now we return to the second integral:
\[ I_2 = \int \frac{2}{(x + 1)^2 + 4} \, dx = 2 \cdot \frac{1}{2} \tan^{-1}\left(\frac{x + 1}{2}\right) = \tan^{-1}\left(\frac{x + 1}{2}\right) \]

### Step 9: Combine All Parts
Finally, we combine \( I_1 \) and \( I_2 \):
\[ I = I_1 + I_2 = \frac{5}{2} \ln |(x + 1)^2 + 4| - \frac{3}{2} \tan^{-1}\left(\frac{x + 1}{2}\right) + \tan^{-1}\left(\frac{x + 1}{2}\right) + C \]
This simplifies to:
\[ I = \frac{5}{2} \ln |(x + 1)^2 + 4| - \frac{1}{2} \tan^{-1}\left(\frac{x + 1}{2}\right) + C \]

### Final Answer
Thus, the final answer is:
\[ I = \frac{5}{2} \ln |(x + 1)^2 + 4| - \frac{1}{2} \tan^{-1}\left(\frac{x + 1}{2}\right) + C \]