`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 will follow these steps:

### Step 1: Simplify the Denominator
We start by rewriting the denominator \( 1 + x + 2x^2 \). To make it easier to integrate, we can complete the square.

\[ 1 + x + 2x^2 = 2\left(x^2 + \frac{x}{2} + \frac{1}{2}\right) \]

Now, we will complete the square for the expression inside the parentheses:

\[ x^2 + \frac{x}{2} + \frac{1}{2} = \left(x + \frac{1}{4}\right)^2 + \frac{1}{4} - \frac{1}{2} = \left(x + \frac{1}{4}\right)^2 + \frac{1}{4} \]

Thus, we have:

\[ 1 + x + 2x^2 = 2\left(\left(x + \frac{1}{4}\right)^2 + \frac{1}{4}\right) \]

### Step 2: Rewrite the Integral
Now, we can rewrite the integral:

\[ I = \int_{0}^{1} \frac{1}{2\left(\left(x + \frac{1}{4}\right)^2 + \frac{1}{4}\right)} \, dx \]

This simplifies to:

\[ I = \frac{1}{2} \int_{0}^{1} \frac{1}{\left(x + \frac{1}{4}\right)^2 + \left(\frac{1}{2}\right)^2} \, dx \]

### Step 3: Use the Arctangent Formula
The integral now resembles the standard form:

\[ \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \]

In our case, \( a = \frac{1}{2} \). Thus, we can apply the formula:

\[ I = \frac{1}{2} \cdot \frac{1}{\frac{1}{2}} \left[ \tan^{-1}\left(\frac{x + \frac{1}{4}}{\frac{1}{2}}\right) \right]_{0}^{1} \]

This simplifies to:

\[ I = \left[ \tan^{-1}\left(2\left(x + \frac{1}{4}\right)\right) \right]_{0}^{1} \]

### Step 4: Evaluate the Limits
Now we evaluate the limits:

1. At \( x = 1 \):

\[ \tan^{-1}\left(2\left(1 + \frac{1}{4}\right)\right) = \tan^{-1}\left(2 \cdot \frac{5}{4}\right) = \tan^{-1}\left(\frac{5}{2}\right) \]

2. At \( x = 0 \):

\[ \tan^{-1}\left(2\left(0 + \frac{1}{4}\right)\right) = \tan^{-1}\left(\frac{1}{2}\right) \]

Thus, we have:

\[ I = \tan^{-1}\left(\frac{5}{2}\right) - \tan^{-1}\left(\frac{1}{2}\right) \]

### Step 5: Final Result
Putting it all together, we get:

\[ I = \tan^{-1}\left(\frac{5}{2}\right) - \tan^{-1}\left(\frac{1}{2}\right) \]