Evaluate : `intsqrt(x+sqrt(x^(2)+2))`dx

Answer

To evaluate the integral \( I = \int \sqrt{x + \sqrt{x^2 + 2}} \, dx \), we will follow these steps:

### Step 1: Simplify the expression inside the integral
We start with the expression \( \sqrt{x + \sqrt{x^2 + 2}} \).

### Step 2: Use a substitution
Let \( u = \sqrt{x + \sqrt{x^2 + 2}} \). Then, we need to express \( x \) in terms of \( u \) and find \( dx \).

### Step 3: Square both sides
Squaring both sides gives:
\[ u^2 = x + \sqrt{x^2 + 2} \]
Rearranging this, we have:
\[ \sqrt{x^2 + 2} = u^2 - x \]
Now, squaring again:
\[ x^2 + 2 = (u^2 - x)^2 \]
Expanding the right side:
\[ x^2 + 2 = u^4 - 2u^2x + x^2 \]
This simplifies to:
\[ 2 = u^4 - 2u^2x \]
Rearranging gives:
\[ 2u^2x = u^4 - 2 \quad \Rightarrow \quad x = \frac{u^4 - 2}{2u^2} \]

### Step 4: Differentiate to find \( dx \)
Now, we differentiate \( x \) with respect to \( u \):
\[ dx = \left( \frac{d}{du} \left( \frac{u^4 - 2}{2u^2} \right) \right) du \]
Using the quotient rule:
\[ dx = \left( \frac{(4u^3)(2u^2) - (u^4 - 2)(4u)}{(2u^2)^2} \right) du \]
This simplifies to:
\[ dx = \frac{8u^5 - 4u^5 + 8u}{4u^4} du = \frac{4u^5 + 8u}{4u^4} du = \left( u + \frac{2}{u^3} \right) du \]

### Step 5: Substitute back into the integral
Now we substitute \( x \) and \( dx \) back into the integral:
\[ I = \int u \left( u + \frac{2}{u^3} \right) du \]
This simplifies to:
\[ I = \int \left( u^2 + \frac{2}{u^2} \right) du \]

### Step 6: Integrate term by term
Now we can integrate term by term:
\[ I = \int u^2 \, du + 2 \int u^{-2} \, du \]
Calculating these integrals gives:
\[ I = \frac{u^3}{3} - \frac{2}{u} + C \]

### Step 7: Substitute back \( u \)
Finally, we substitute back \( u = \sqrt{x + \sqrt{x^2 + 2}} \):
\[ I = \frac{(x + \sqrt{x^2 + 2})^{3/2}}{3} - \frac{2}{\sqrt{x + \sqrt{x^2 + 2}}} + C \]

### Final Result:
Thus, the evaluated integral is:
\[ I = \frac{(x + \sqrt{x^2 + 2})^{3/2}}{3} - \frac{2}{\sqrt{x + \sqrt{x^2 + 2}}} + C \]