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

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) \(\frac{\pi}{8} \log 2\).