Choose the correct answer
`intsqrt(1+x^2)dx` is equal to

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 use a trigonometric substitution. Here are the steps to arrive at the solution:

### Step 1: Trigonometric Substitution
Let \(x = \tan(\theta)\). Then, we have:
\[ dx = \sec^2(\theta) \, d\theta \]
And, using the identity \(1 + \tan^2(\theta) = \sec^2(\theta)\), we can rewrite the integral:
\[ \sqrt{1+x^2} = \sqrt{1+\tan^2(\theta)} = \sec(\theta) \]

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

### Step 3: Integrate \(\sec^3(\theta)\)
The integral of \(\sec^3(\theta)\) can be computed using the formula:
\[ \int \sec^3(\theta) \, d\theta = \frac{1}{2} \sec(\theta) \tan(\theta) + \frac{1}{2} \ln |\sec(\theta) + \tan(\theta)| + C \]

### Step 4: Substitute Back
Now we need to substitute back in terms of \(x\):
- Since \(x = \tan(\theta)\), we have \(\tan(\theta) = x\) and \(\sec(\theta) = \sqrt{1+x^2}\).
Thus:
\[ \sec(\theta) \tan(\theta) = \sqrt{1+x^2} \cdot x \]
And:
\[ \sec(\theta) + \tan(\theta) = \sqrt{1+x^2} + x \]

### Step 5: Final Result
Putting everything together, we get:
\[ \int \sqrt{1+x^2} \, dx = \frac{1}{2} x \sqrt{1+x^2} + \frac{1}{2} \ln |\sqrt{1+x^2} + x| + C \]

### Conclusion
Thus, the final answer is:
\[ \int \sqrt{1+x^2} \, dx = \frac{1}{2} x \sqrt{1+x^2} + \frac{1}{2} \ln |\sqrt{1+x^2} + x| + C \]