`int(sqrt(x^2+10 x+24))/(x+5)dx` is equal to

Options

A. `sqrt(x^(2)+10x+24)+sec^(-1)(x+5)+c`
B. `sqrt(x^(2)+10x+24)-"cosec"^(-1)(x+5)+c`
C. `sec^(-1)(x+5)-sqrt(x^(2)+10x+24)+c`
D. `sqrt(x^(2)+10x+24)-sec^(-1)(x+5)+c`

Explanation

To solve the integral \( \int \frac{\sqrt{x^2 + 10x + 24}}{x + 5} \, dx \), we can follow these steps:

### Step 1: Simplify the Expression Under the Square Root
First, we simplify the expression under the square root:
\[ x^2 + 10x + 24 = (x + 5)^2 - 1 \]
This means we can rewrite the integral as:
\[ \int \frac{\sqrt{(x + 5)^2 - 1}}{x + 5} \, dx \]

### Step 2: Use Trigonometric Substitution
Next, we can use the substitution \( x + 5 = \sec(\theta) \). Therefore, \( dx = \sec(\theta) \tan(\theta) \, d\theta \). The integral now becomes:
\[ \int \frac{\sqrt{\sec^2(\theta) - 1}}{\sec(\theta)} \sec(\theta) \tan(\theta) \, d\theta \]
Since \( \sqrt{\sec^2(\theta) - 1} = \tan(\theta) \), we can simplify the integral to:
\[ \int \tan^2(\theta) \, d\theta \]

### Step 3: Integrate \( \tan^2(\theta) \)
Using the identity \( \tan^2(\theta) = \sec^2(\theta) - 1 \), we can rewrite the integral:
\[ \int \tan^2(\theta) \, d\theta = \int (\sec^2(\theta) - 1) \, d\theta = \int \sec^2(\theta) \, d\theta - \int 1 \, d\theta \]
The integrals evaluate to:
\[ \tan(\theta) - \theta + C \]

### Step 4: Substitute Back to Original Variable
Now we need to substitute back to the variable \( x \). Recall that:
\[ \tan(\theta) = \sqrt{(x + 5)^2 - 1} = \sqrt{x^2 + 10x + 24 - 1} = \sqrt{x^2 + 10x + 23} \]
And since \( \theta = \sec^{-1}(x + 5) \), we have:
\[ \int \frac{\sqrt{x^2 + 10x + 24}}{x + 5} \, dx = \sqrt{x^2 + 10x + 23} - \sec^{-1}(x + 5) + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{\sqrt{x^2 + 10x + 24}}{x + 5} \, dx = \sqrt{x^2 + 10x + 23} - \sec^{-1}(x + 5) + C \]