If `int (x+sqrt(1+x^(2)))^(n)dx`.
`=(1)/(a(n+1)){x+sqrt(1+x^(2))}^(n+1)+(1)/(-b(n-1)){x+sqrt(1+x^(2))}^(n-1)+C` Then `(a+b)` is equal to

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the integral \( \int (x + \sqrt{1 + x^2})^n \, dx \) and express it in the given form, we will follow these steps:

### Step 1: Substitution
Let:
\[ t = x + \sqrt{1 + x^2} \]
Then, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = 1 + \frac{x}{\sqrt{1 + x^2}} \]
This simplifies to:
\[ \frac{dt}{dx} = \frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2}} = \frac{t}{\sqrt{1 + x^2}} \]
Thus, we can express \( dx \) as:
\[ dx = \frac{\sqrt{1 + x^2}}{t} \, dt \]

### Step 2: Express \( \sqrt{1 + x^2} \) in terms of \( t \)
From the substitution \( t = x + \sqrt{1 + x^2} \), we can isolate \( \sqrt{1 + x^2} \):
\[ \sqrt{1 + x^2} = t - x \]
Squaring both sides gives:
\[ 1 + x^2 = (t - x)^2 \]
Expanding and rearranging:
\[ 1 + x^2 = t^2 - 2tx + x^2 \]
This simplifies to:
\[ 1 = t^2 - 2tx \quad \Rightarrow \quad 2tx = t^2 - 1 \quad \Rightarrow \quad x = \frac{t^2 - 1}{2t} \]

### Step 3: Substitute back into the integral
Now, we need to express \( \sqrt{1 + x^2} \) in terms of \( t \):
\[ \sqrt{1 + x^2} = \sqrt{1 + \left(\frac{t^2 - 1}{2t}\right)^2} \]
Calculating this gives:
\[ \sqrt{1 + x^2} = \sqrt{\frac{4t^2 + (t^2 - 1)^2}{4t^2}} = \frac{\sqrt{t^4 + 2t^2 + 1}}{2t} = \frac{(t^2 + 1)}{2t} \]

### Step 4: Substitute \( dx \) and simplify the integral
Now substituting \( dx \) and \( \sqrt{1 + x^2} \) back into the integral:
\[ \int (t)^n \cdot \frac{(t^2 + 1)}{2t} \, dt = \frac{1}{2} \int (t^{n-1}(t^2 + 1)) \, dt \]
This can be split into two integrals:
\[ \frac{1}{2} \left( \int t^{n+1} \, dt + \int t^{n-1} \, dt \right) \]

### Step 5: Integrate
Calculating these integrals:
\[ \frac{1}{2} \left( \frac{t^{n+1}}{n+1} + \frac{t^{n-1}}{n-1} \right) + C \]

### Step 6: Back-substitute \( t \)
Now substituting back \( t = x + \sqrt{1 + x^2} \):
\[ = \frac{1}{2(n+1)}(x + \sqrt{1 + x^2})^{n+1} + \frac{1}{2(-b)(n-1)}(x + \sqrt{1 + x^2})^{n-1} + C \]

### Step 7: Identify constants \( a \) and \( b \)
From the expression, we can identify:
- \( a = 2 \)
- \( b = 2 \)

Thus, \( a + b = 2 + 2 = 4 \).

### Final Answer
\[ \boxed{4} \]