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 use a substitution and properties of definite integrals. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( x = \tan \theta \). Then, we have:
\[ dx = \sec^2 \theta \, d\theta \]
When \( x = 0 \), \( \theta = 0 \) and when \( x = 1 \), \( \theta = \frac{\pi}{4} \).

### Step 2: Change of Limits and Expression
Now substituting in the integral, we get:
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{8 \log(1 + \tan \theta)}{1 + \tan^2 \theta} \sec^2 \theta \, d\theta \]
Since \( 1 + \tan^2 \theta = \sec^2 \theta \), the integral simplifies to:
\[ I = \int_{0}^{\frac{\pi}{4}} 8 \log(1 + \tan \theta) \, d\theta \]

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

### Step 4: Simplifying the Logarithm
This simplifies to:
\[ I = 8 \int_{0}^{\frac{\pi}{4}} \log\left(\frac{2}{1 + \tan \theta}\right) \, d\theta \]
Breaking it down using properties of logarithms:
\[ I = 8 \int_{0}^{\frac{\pi}{4}} \log(2) \, d\theta - 8 \int_{0}^{\frac{\pi}{4}} \log(1 + \tan \theta) \, d\theta \]

### Step 5: Combining the Integrals
Let \( I = 8 \int_{0}^{\frac{\pi}{4}} \log(2) \, d\theta - 8I \):
\[ I = 8 \log(2) \cdot \frac{\pi}{4} - 8I \]
\[ 9I = 2\pi \log(2) \]
\[ I = \frac{2\pi \log(2)}{9} \]

### Final Result
Thus, the value of the integral is:
\[ I = \frac{2\pi \log(2)}{9} \]