int_0^1 (tan^-1 x)^2/(1+x^2) dx =?

ধরি, \( I = \int_0^1 \frac{(\tan^{-1} x)^2}{1+x^2} dx \)

ধরি, \( x = \tan \theta \), সুতরাং \( dx = \sec^2 \theta d\theta \)

যখন \( x = 0 \), তখন \( \theta = 0 \) এবং যখন \( x = 1 \), তখন \( \theta = \frac{\pi}{4} \)

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

\( = \int_0^{\pi/4} \theta^2 d\theta \)

\( = \left[ \frac{\theta^3}{3} \right]_0^{\pi/4} \)

\( = \frac{1}{3} \left( \frac{\pi}{4} \right)^3 - \frac{1}{3}(0)^3 \)

\( = \frac{1}{3} \cdot \frac{\pi^3}{64} \)

\( = \frac{\pi^3}{192} \)

অতএব, \( \int_0^1 \frac{(\tan^{-1} x)^2}{1+x^2} dx = \frac{\pi^3}{192} \) 🥳