`Evaluate : int1/(x^2+2x+5)dx`

Answer

To evaluate the integral \( \int \frac{1}{x^2 + 2x + 5} \, dx \), we will follow these steps:

### Step 1: Rewrite the quadratic expression
First, we need to rewrite the expression \( x^2 + 2x + 5 \) in a form that is easier to integrate. We can complete the square for the quadratic part:

\[ x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x + 1)^2 + 4 \]

### Step 2: Substitute the expression into the integral
Now, we can substitute this back into the integral:

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

### Step 3: Identify the standard form
The integral now resembles the standard form \( \int \frac{1}{x^2 + a^2} \, dx \), where \( a^2 = 4 \) (thus \( a = 2 \)). The formula for this integral is:

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

### Step 4: Apply the formula
In our case, we have \( x \) replaced by \( x + 1 \) and \( a = 2 \):

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

### Step 5: Write the final answer
Thus, the final answer to the integral is:

\[ \int \frac{1}{x^2 + 2x + 5} \, dx = \frac{1}{2} \tan^{-1} \left( \frac{x + 1}{2} \right) + C \]