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 evaluate the integral

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

we will use the substitution \( x = \tan \theta \).

### Step 1: Change of Variables

Let \( x = \tan \theta \). Then, we have:

\[ dx = \sec^2 \theta \, d\theta. \]

Now, we need to change the limits of integration. When \( x = 0 \), \( \theta = 0 \) and when \( x = 1 \), \( \theta = \frac{\pi}{4} \).

### Step 2: Rewrite the Integral

Substituting \( x = \tan \theta \) into the integral, we get:

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

Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \), the integral simplifies to:

\[ I = \int_{0}^{\frac{\pi}{4}} \log(1+\tan \theta) \, d\theta. \]

### Step 3: Symmetry Property of the Integral

Now, we will use the property of integrals:

\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx. \]

In our case, we will apply this property with \( a = \frac{\pi}{4} \):

\[ I = \int_{0}^{\frac{\pi}{4}} \log(1+\tan(\frac{\pi}{4} - \theta)) \, d\theta. \]

Using the identity \( \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan \theta}{1 + \tan \theta} \), we have:

\[ I = \int_{0}^{\frac{\pi}{4}} \log\left(1 + \frac{1 - \tan \theta}{1 + \tan \theta}\right) \, d\theta. \]

### Step 4: Simplifying the Logarithm

Now, simplify the logarithm:

\[ 1 + \frac{1 - \tan \theta}{1 + \tan \theta} = \frac{(1 + \tan \theta) + (1 - \tan \theta)}{1 + \tan \theta} = \frac{2}{1 + \tan \theta}. \]

Thus, we can rewrite the integral as:

\[ I = \int_{0}^{\frac{\pi}{4}} \log\left(\frac{2}{1+\tan \theta}\right) \, d\theta. \]

### Step 5: Separate the Integral

Now, we can separate the integral:

\[ I = \int_{0}^{\frac{\pi}{4}} \log 2 \, d\theta - \int_{0}^{\frac{\pi}{4}} \log(1+\tan \theta) \, d\theta. \]

The first integral evaluates to:

\[ \log 2 \cdot \frac{\pi}{4}. \]

Thus, we have:

\[ I = \frac{\pi}{4} \log 2 - I. \]

### Step 6: Solve for \( I \)

Now, add \( I \) to both sides:

\[ 2I = \frac{\pi}{4} \log 2. \]

Dividing by 2 gives:

\[ I = \frac{\pi}{8} \log 2. \]

### Conclusion

Therefore, the value of the integral is:

\[ \int_{0}^{1} \frac{\log(1+x)}{1+x^2} \, dx = \frac{\pi}{8} \log 2. \]