Evaluate: `int (sec^2x)/sqrt(tan^2x+2tanx+5) dx`

Answer

To evaluate the integral

\[ I = \int \frac{\sec^2 x}{\sqrt{\tan^2 x + 2\tan x + 5}} \, dx, \]

we will follow these steps:

### Step 1: Substitution
Let \( u = \tan x \). Then, we have:

\[ du = \sec^2 x \, dx. \]

This means that \( dx = \frac{du}{\sec^2 x} \).

### Step 2: Rewrite the Integral
Substituting \( u \) into the integral, we get:

\[ I = \int \frac{1}{\sqrt{u^2 + 2u + 5}} \, du. \]

### Step 3: Simplify the Expression Inside the Square Root
We can simplify the expression under the square root:

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

### Step 4: Substitute Back into the Integral
Now, we rewrite the integral:

\[ I = \int \frac{1}{\sqrt{(u + 1)^2 + 4}} \, du. \]

### Step 5: Use a Standard Integral Formula
This integral can be solved using the standard integral formula:

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

In our case, \( x = u + 1 \) and \( a = 2 \). Thus, we have:

\[ I = \ln |(u + 1) + \sqrt{(u + 1)^2 + 4}| + C. \]

### Step 6: Substitute Back for \( u \)
Now, substituting back \( u = \tan x \):

\[ I = \ln |\tan x + 1 + \sqrt{(\tan x + 1)^2 + 4}| + C. \]

### Final Answer
Thus, the final answer is:

\[ I = \ln |\tan x + 1 + \sqrt{\tan^2 x + 2\tan x + 5}| + C. \]