`int_(0)^(1) (1)/(1+x+2x^(2))dx`

Answer

To solve the integral \( I = \int_{0}^{1} \frac{1}{1 + x + 2x^2} \, dx \), we can follow these steps:

### Step 1: Simplify the Integral
We start by factoring out a 2 from the denominator:
\[ I = \int_{0}^{1} \frac{1}{1 + x + 2x^2} \, dx = \frac{1}{2} \int_{0}^{1} \frac{1}{\frac{1}{2} + \frac{x}{2} + x^2} \, dx \]

### Step 2: Complete the Square
Next, we rewrite the quadratic in the denominator by completing the square:
\[ \frac{1}{2} + \frac{x}{2} + x^2 = x^2 + \frac{x}{2} + \frac{1}{2} \]
To complete the square, we take half of the coefficient of \( x \) (which is \( \frac{1}{2} \)), square it (getting \( \frac{1}{16} \)), and adjust the expression:
\[ x^2 + \frac{x}{2} + \frac{1}{2} = \left(x + \frac{1}{4}\right)^2 + \left(\frac{1}{2} - \frac{1}{16}\right) = \left(x + \frac{1}{4}\right)^2 + \frac{7}{16} \]

### Step 3: Substitute and Change Limits
Now we substitute \( t = x + \frac{1}{4} \), which gives \( dx = dt \) and changes the limits:
- When \( x = 0 \), \( t = \frac{1}{4} \)
- When \( x = 1 \), \( t = \frac{5}{4} \)

Thus, the integral becomes:
\[ I = \frac{1}{2} \int_{\frac{1}{4}}^{\frac{5}{4}} \frac{1}{\left(t\right)^2 + \frac{7}{16}} \, dt \]

### Step 4: Use the Integral Formula
We can use the formula for the integral of the form \( \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C \). Here, \( a^2 = \frac{7}{16} \) implies \( a = \frac{\sqrt{7}}{4} \):
\[ I = \frac{1}{2} \cdot \frac{4}{\sqrt{7}} \left[ \tan^{-1} \left(\frac{t}{\frac{\sqrt{7}}{4}}\right) \right]_{\frac{1}{4}}^{\frac{5}{4}} \]

### Step 5: Evaluate the Limits
Now we evaluate the integral at the limits:
\[ I = \frac{2}{\sqrt{7}} \left[ \tan^{-1} \left(\frac{5}{\sqrt{7}}\right) - \tan^{-1} \left(\frac{1}{\sqrt{7}}\right) \right] \]

### Final Result
Thus, the final result for the integral is:
\[ I = \frac{2}{\sqrt{7}} \left[ \tan^{-1} \left(\frac{5}{\sqrt{7}}\right) - \tan^{-1} \left(\frac{1}{\sqrt{7}}\right) \right] \]