`int_(1)^(2) (x)/(sqrt(1+2x^(2)))dx`

Answer

To solve the integral

\[ I = \int_{1}^{2} \frac{x}{\sqrt{1 + 2x^2}} \, dx, \]

we will follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral:

\[ I = \int_{1}^{2} \frac{x}{\sqrt{1 + 2x^2}} \, dx. \]

### Step 2: Multiply and Divide by 4
Next, we multiply and divide the integrand by 4:

\[ I = \int_{1}^{2} \frac{4x}{4\sqrt{1 + 2x^2}} \, dx = \int_{1}^{2} \frac{4x}{\sqrt{4(1 + 2x^2)}} \, dx. \]

### Step 3: Substitution
Let us make the substitution \( t = 1 + 2x^2 \). Then, we differentiate:

\[ dt = 4x \, dx \quad \Rightarrow \quad dx = \frac{dt}{4x}. \]

### Step 4: Change the Limits
Now we need to change the limits of integration. When \( x = 1 \):

\[ t = 1 + 2(1^2) = 3, \]

and when \( x = 2 \):

\[ t = 1 + 2(2^2) = 9. \]

### Step 5: Substitute in the Integral
Now substituting \( t \) and \( dx \) into the integral, we have:

\[ I = \int_{3}^{9} \frac{1}{\sqrt{t}} \, dt. \]

### Step 6: Integrate
The integral of \( \frac{1}{\sqrt{t}} \) is:

\[ \int \frac{1}{\sqrt{t}} \, dt = 2\sqrt{t}. \]

Thus,

\[ I = 2\sqrt{t} \bigg|_{3}^{9} = 2\sqrt{9} - 2\sqrt{3} = 6 - 2\sqrt{3}. \]

### Step 7: Final Result
Thus, we have:

\[ I = 6 - 2\sqrt{3}. \]

### Step 8: Simplify
We can also express this as:

\[ I = 2(3 - \sqrt{3}). \]

### Conclusion
The final value of the integral is:

\[ I = 2(3 - \sqrt{3}). \]