int tan^-1x/(1+x^2)=?
সঠিক উত্তরঃ
D.
1/2(tan^-1x)^2+k
Explanation: 
Another Explanation (5):
সমাধান:
ধরি, \(tan^{-1}x = z\) 🤩
তাহলে, \(\frac{1}{1+x^2} dx = dz\) 😎
সুতরাং, \(\int \frac{tan^{-1}x}{1+x^2} dx = \int z dz\) 🥳
\(= \frac{z^2}{2} + k\) 🙏
\(= \frac{(tan^{-1}x)^2}{2} + k\) 🤓
\(= \frac{1}{2}(tan^{-1}x)^2 + k\) 🎉
সুতরাং, \(\int \frac{tan^{-1}x}{1+x^2} dx = \frac{1}{2}(tan^{-1}x)^2 + k\) 💖