Question

The value of \(\int\limits_0^1\frac{\log(1+x)}{1+x^2}\ dx\) is

• A. \(\frac{\pi}{2}\log2\)
• B. \(\frac{\pi}{4}\log2\)
• C. \(\frac{1}{2}\)
• D. \(\frac{\pi}{8}\log2\)

Answer 1

The Correct Option is D

Let:

$$ I = \int_0^1 \frac{\log(1+x)}{1+x^2} dx. $$

Let $ x = \tan \theta $. 

Then $ dx = \sec^2 \theta d\theta $ and $ 1+x^2 = 1+\tan^2 \theta = \sec^2 \theta $. When $ x=0 $, $ \theta = 0 $. When $ x=1 $, $ \theta = \frac{\pi}{4} $. 

Substitute into the integral:

$$ I = \int_0^{\pi/4} \frac{\log(1+\tan \theta)}{\sec^2 \theta} \sec^2 \theta d\theta = \int_0^{\pi/4} \log(1+\tan \theta) d\theta. $$

We use the property $ \int_a^b f(x) dx = \int_a^b f(a+b-x) dx $. 

Applying this to $ I $:

$$ I = \int_0^{\pi/4} \log(1+\tan (\frac{\pi}{4} - \theta)) d\theta. $$

Using the tangent subtraction formula $ \tan(\frac{\pi}{4} - \theta) = \frac{\tan \frac{\pi}{4} - \tan \theta}{1 + \tan \frac{\pi}{4} \tan \theta} = \frac{1 - \tan \theta}{1 + \tan \theta} $, we get:

$$ I = \int_0^{\pi/4} \log\left(1 + \frac{1 - \tan \theta}{1 + \tan \theta}\right) d\theta. $$

Simplify the argument of 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:

$$ I = \int_0^{\pi/4} \log\left(\frac{2}{1 + \tan \theta}\right) d\theta = \int_0^{\pi/4} \left(\log 2 - \log(1+\tan \theta)\right) d\theta. $$

Split the integral:

$$ I = \int_0^{\pi/4} \log 2 d\theta - \int_0^{\pi/4} \log(1+\tan \theta) d\theta. $$

The first term evaluates as:

$$ \int_0^{\pi/4} \log 2 d\theta = \log 2 \cdot \theta \Big|_0^{\pi/4} = \log 2 \cdot \left(\frac{\pi}{4} - 0\right) = \frac{\pi}{4} \log 2. $$

The second term is just $ I $, so:

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

Rearrange to solve for $ I $:

$$ 2I = \frac{\pi}{4} \log 2 \implies I = \frac{\pi}{8} \log 2. $$

Answer 2

We need to find the value of \(\int_{0}^{1} \frac{\log(1+x)}{1+x^2} \, dx\).

Let \(x = \tan \theta\). Then \(dx = \sec^2 \theta \, d\theta\).

When \(x = 0\), \(\tan \theta = 0\), so \(\theta = 0\).

When \(x = 1\), \(\tan \theta = 1\), so \(\theta = \frac{\pi}{4}\).

So the integral becomes:

\(\int_{0}^{\pi/4} \frac{\log(1+\tan \theta)}{1+\tan^2 \theta} \sec^2 \theta \, d\theta\)

Since \(1 + \tan^2 \theta = \sec^2 \theta\), we have:

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

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

Using the property \(\int_a^b f(x) dx = \int_a^b f(a+b-x) dx\), we have:

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

Using the formula for \(\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\), we get:

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

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

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

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

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

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

\(I = \log 2 [\theta]_{0}^{\pi/4} - I\)

\(I = \log 2 \cdot \frac{\pi}{4} - I\)

\(2I = \frac{\pi}{4} \log 2\)

\(I = \frac{\pi}{8} \log 2\)

Therefore, the correct option is (D) \(\frac{\pi}{8} \log 2\).