The value of integral`int_(0)^(1)(log(1+x))/(1+x^(2))dx`, is

Options

A. `(pi)/(8)log_(e )2`
B. `(pi)/(4)log_(e )2`
C. `-(pi)/(8)log_(e )2`
D. `-(pi)/(4) log_(e ) 2`

Explanation

To solve the integral \( I = \int_{0}^{1} \frac{\log(1+x)}{1+x^2} \, dx \), we will follow a systematic approach:

### Step 1: Substitution
We will use the substitution \( x = \tan \theta \). This gives us \( dx = \sec^2 \theta \, d\theta \). The limits change as follows:
- When \( x = 0 \), \( \theta = 0 \)
- When \( x = 1 \), \( \theta = \frac{\pi}{4} \)

Thus, the integral becomes:
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\log(1+\tan \theta)}{1+\tan^2 \theta} \sec^2 \theta \, d\theta \]

### Step 2: Simplifying the Integral
Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \), we can simplify the integral:
\[ I = \int_{0}^{\frac{\pi}{4}} \log(1+\tan \theta) \, d\theta \]

### Step 3: Using the Property of Integrals
We can use the property of integrals:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \]
In our case, \( a = \frac{\pi}{4} \). Therefore, we have:
\[ I = \int_{0}^{\frac{\pi}{4}} \log\left(1 + \tan\left(\frac{\pi}{4} - \theta\right)\right) \, d\theta \]

### Step 4: Simplifying Further
Using the identity \( \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan \theta}{1 + \tan \theta} \), we can rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{4}} \log\left(1 + \frac{1 - \tan \theta}{1 + \tan \theta}\right) \, d\theta \]
This simplifies to:
\[ I = \int_{0}^{\frac{\pi}{4}} \log\left(\frac{2}{1 + \tan \theta}\right) \, d\theta \]

### Step 5: Splitting the Logarithm
We can split the logarithm:
\[ I = \int_{0}^{\frac{\pi}{4}} \log(2) \, d\theta - \int_{0}^{\frac{\pi}{4}} \log(1 + \tan \theta) \, d\theta \]
The first integral evaluates to:
\[ \log(2) \cdot \frac{\pi}{4} \]

### Step 6: Combining the Results
Now we have:
\[ I = \frac{\pi}{4} \log(2) - I \]
Adding \( I \) to both sides gives:
\[ 2I = \frac{\pi}{4} \log(2) \]
Thus,
\[ I = \frac{\pi}{8} \log(2) \]

### Final Answer
The value of the integral is:
\[ \int_{0}^{1} \frac{\log(1+x)}{1+x^2} \, dx = \frac{\pi}{8} \log(2) \]