int_0^1 tan^-1x/(1+x^2) dx=? 

ধরি, \( I = \int_{0}^{1} \frac{\tan^{-1}x}{1+x^2} dx \) 🧐

এখন, \( \tan^{-1}x = \theta \) ধরি। সুতরাং, \( x = \tan\theta \)।

অতএব, \( dx = \sec^2\theta d\theta = (1+\tan^2\theta) d\theta = (1+x^2)d\theta \).

যখন \( x = 0 \), তখন \( \theta = \tan^{-1}(0) = 0 \)।

যখন \( x = 1 \), তখন \( \theta = \tan^{-1}(1) = \frac{\pi}{4} \).

সুতরাং, আমাদের ইন্টিগ্রালটি হবে:

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

\( = \int_{0}^{\frac{\pi}{4}} \theta d\theta \)

\( = \left[ \frac{\theta^2}{2} \right]_{0}^{\frac{\pi}{4}} \)

\( = \frac{1}{2} \left[ \left(\frac{\pi}{4}\right)^2 - 0^2 \right] \)

\( = \frac{1}{2} \cdot \frac{\pi^2}{16} \)

\( = \frac{\pi^2}{32} \)

সুতরাং, \( \int_{0}^{1} \frac{\tan^{-1}x}{1+x^2} dx = \frac{\pi^2}{32} \) 🎉