The value of `int_(0)^(1)(8log(1+x))/(1+x^(2))dx` is

Options

A. `log 2`
B. `pi log 2`
C. `(pi)/8 log 2`
D. `(pi)/2 log 2`

Explanation

To solve the integral \( I = \int_{0}^{1} \frac{8 \log(1+x)}{1+x^2} \, dx \), we will follow these steps:

### Step 1: Substitution
We will use the substitution \( x = \tan \theta \). Then, \( dx = \sec^2 \theta \, d\theta \). We also need to change the limits of integration:
- When \( x = 0 \), \( \theta = \tan^{-1}(0) = 0 \).
- When \( x = 1 \), \( \theta = \tan^{-1}(1) = \frac{\pi}{4} \).

Thus, the integral becomes:
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{8 \log(1 + \tan \theta)}{1 + \tan^2 \theta} \sec^2 \theta \, d\theta \]

### Step 2: Simplifying the Integral
Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \), we can simplify the integral:
\[ I = \int_{0}^{\frac{\pi}{4}} 8 \log(1 + \tan \theta) \, d\theta \]

### Step 3: Using Symmetry Property
We can use the property of integrals:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, we can express \( I \) as:
\[ I = \int_{0}^{\frac{\pi}{4}} 8 \log(1 + \tan(\frac{\pi}{4} - \theta)) \, d\theta \]
Using the identity \( \tan(\frac{\pi}{4} - \theta) = \frac{1 - \tan \theta}{1 + \tan \theta} \), we find:
\[ I = \int_{0}^{\frac{\pi}{4}} 8 \log\left(1 + \frac{1 - \tan \theta}{1 + \tan \theta}\right) \, d\theta \]

### Step 4: Simplifying the Logarithm
This simplifies to:
\[ I = \int_{0}^{\frac{\pi}{4}} 8 \log\left(\frac{2}{1 + \tan \theta}\right) \, d\theta \]
This can be separated into two integrals:
\[ I = \int_{0}^{\frac{\pi}{4}} 8 \log(2) \, d\theta - \int_{0}^{\frac{\pi}{4}} 8 \log(1 + \tan \theta) \, d\theta \]
Let \( J = \int_{0}^{\frac{\pi}{4}} 8 \log(1 + \tan \theta) \, d\theta \). Then:
\[ I = 8 \log(2) \cdot \frac{\pi}{4} - J \]
Since \( I = J \), we can write:
\[ I = 8 \log(2) \cdot \frac{\pi}{4} - I \]

### Step 5: Solving for I
Adding \( I \) to both sides gives:
\[ 2I = 8 \log(2) \cdot \frac{\pi}{4} \]
Thus,
\[ I = 4 \log(2) \cdot \frac{\pi}{4} = \pi \log(2) \]

### Final Answer
The value of the integral is:
\[ \boxed{\pi \log(2)} \]