`int (1)/((x+5)sqrt(x+4))dx` is :

Options

A. `tan^(-1) sqrt(x+4)+C`
B. `2tan^(-1) sqrt(x+4)+C`
C. `-tan^(-1) sqrt(x+4) + C`
D. `-2 tan^(-1)sqrt(x+4) + C`

Answer

To solve the integral \( I = \int \frac{1}{(x+5)\sqrt{x+4}} \, dx \), we will follow these steps:

### Step 1: Simplify the Integrand
We can rewrite \( x + 5 \) as \( (x + 4) + 1 \). Therefore, we can express the integrand as:
\[ I = \int \frac{1}{(x + 4 + 1)\sqrt{x + 4}} \, dx = \int \frac{1}{(x + 4)\sqrt{x + 4} + \sqrt{x + 4}} \, dx \]

### Step 2: Substitution
Let \( t = \sqrt{x + 4} \). Then, we have:
\[ x + 4 = t^2 \quad \Rightarrow \quad x = t^2 - 4 \]
Differentiating both sides gives:
\[ dx = 2t \, dt \]

### Step 3: Substitute in the Integral
Now, substitute \( t \) and \( dx \) into the integral:
\[ I = \int \frac{1}{(t^2 - 4 + 5)t} \cdot 2t \, dt = \int \frac{2t}{(t^2 + 1)t} \, dt = \int \frac{2}{t^2 + 1} \, dt \]

### Step 4: Integrate
The integral \( \int \frac{2}{t^2 + 1} \, dt \) can be computed as:
\[ I = 2 \tan^{-1}(t) + C \]

### Step 5: Back Substitute
Now, substitute back \( t = \sqrt{x + 4} \):
\[ I = 2 \tan^{-1}(\sqrt{x + 4}) + C \]

### Final Answer
Thus, the integral evaluates to:
\[ \int \frac{1}{(x+5)\sqrt{x+4}} \, dx = 2 \tan^{-1}(\sqrt{x + 4}) + C \]

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