Choose the correct answer`intsqrt(1+x^2)dx`is equal to(A) `x/2sqrt(1+x^2)+1/2log|(x+sqrt(x+x^2))|+C`(B) `2/3(1+x^2)^(3/2)+C` (C) `2/3x(1+x^2)^(3/2)+C` (D) `(x^2)/2sqrt(1+x^2)+1/2x^2log|x+sqrt(1+x^2)|+C`

Options

A. `x/2sqrt(1+x^2)+1/2log|(x+sqrt(x+x^2))|+C`
B. `2/3(1+x^2)^(3/2)+C`
C. `2/3x(1+x^2)^(3/2)+C`
D. `(x^2)/2sqrt(1+x^2)+1/2x^2log|x+sqrt(1+x^2)|+C`

Explanation

To solve the integral \( \int \sqrt{1+x^2} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( x = \tan \theta \). Then, differentiating both sides gives:
\[ dx = \sec^2 \theta \, d\theta \]
Also, we know that:
\[ 1 + \tan^2 \theta = \sec^2 \theta \quad \Rightarrow \quad \sqrt{1 + x^2} = \sqrt{1 + \tan^2 \theta} = \sec \theta \]

### Step 2: Substitute in the integral
Substituting \( x \) and \( dx \) into the integral gives:
\[ \int \sqrt{1+x^2} \, dx = \int \sec \theta \cdot \sec^2 \theta \, d\theta = \int \sec^3 \theta \, d\theta \]

### Step 3: Integrate \( \sec^3 \theta \)
To integrate \( \sec^3 \theta \), we can use integration by parts. Let:
- \( u = \sec \theta \)
- \( dv = \sec^2 \theta \, d\theta \)

Then, we have:
\[ du = \sec \theta \tan \theta \, d\theta \quad \text{and} \quad v = \tan \theta \]
Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
This gives:
\[ \int \sec^3 \theta \, d\theta = \sec \theta \tan \theta - \int \tan \theta \sec \theta \tan \theta \, d\theta \]
The integral \( \int \tan^2 \theta \sec \theta \, d\theta \) can be simplified using \( \tan^2 \theta = \sec^2 \theta - 1 \):
\[ \int \tan^2 \theta \sec \theta \, d\theta = \int (\sec^2 \theta - 1) \sec \theta \, d\theta = \int \sec^3 \theta \, d\theta - \int \sec \theta \, d\theta \]
Let \( I = \int \sec^3 \theta \, d\theta \). Then:
\[ I = \sec \theta \tan \theta - (I - \int \sec \theta \, d\theta) \]
This simplifies to:
\[ 2I = \sec \theta \tan \theta + \int \sec \theta \, d\theta \]
Thus:
\[ I = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \int \sec \theta \, d\theta \]

### Step 4: Evaluate \( \int \sec \theta \, d\theta \)
The integral \( \int \sec \theta \, d\theta \) is known to be:
\[ \int \sec \theta \, d\theta = \ln |\sec \theta + \tan \theta| + C \]

### Step 5: Substitute back to \( x \)
Now substituting back:
\[ I = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln |\sec \theta + \tan \theta| + C \]
Recall that:
\[ \sec \theta = \sqrt{1+x^2}, \quad \tan \theta = x \]
Thus:
\[ I = \frac{1}{2} x \sqrt{1+x^2} + \frac{1}{2} \ln |x + \sqrt{1+x^2}| + C \]

### Final Answer
So, the final result is:
\[ \int \sqrt{1+x^2} \, dx = \frac{x}{2} \sqrt{1+x^2} + \frac{1}{2} \ln |x + \sqrt{1+x^2}| + C \]

### Conclusion
The correct option is (A):
\[ \frac{x}{2} \sqrt{1+x^2} + \frac{1}{2} \ln |x + \sqrt{1+x^2}| + C \]