Explanation: 
Another Explanation (5):
সমাধান:
ধরি, \(I = \int \frac{2\tan^{-1}x}{1+x^2} dx\)
এখন, \(\tan^{-1}x = t\) ধরলে,
\(\frac{1}{1+x^2} dx = dt\) হবে।
সুতরাং, \(I = \int 2t dt\)
\(I = 2 \int t dt\)
\(I = 2 \cdot \frac{t^2}{2} + c\)
\(I = t^2 + c\)
\(I = (\tan^{-1}x)^2 + c\)
অতএব, \(f(x) = (\tan^{-1}x)^2\)
সুতরাং, উত্তর: \((\tan^{-1}x)^2\) 🎉