int(sec^2(cot^-1x))/(1+x^2) dx = কত?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(cot^{-1}x = \theta\) সুতরাং, \(x = cot \theta\) অতএব, \(dx = -cosec^2 \theta d\theta\) এখন, \(\int \frac{sec^2(cot^{-1}x)}{1+x^2} dx = \int \frac{sec^2 \theta}{1+cot^2 \theta} (-cosec^2 \theta) d\theta\) \( = \int \frac{sec^2 \theta}{cosec^2 \theta} (-cosec^2 \theta) d\theta\) \( = - \int sec^2 \theta d\theta\) \( = -tan \theta + c\) যেহেতু \(cot \theta = x\), সুতরাং \(tan \theta = \frac{1}{x}\) অতএব, \(\int \frac{sec^2(cot^{-1}x)}{1+x^2} dx = -\frac{1}{x} + c\) 🥳 সুতরাং, উত্তর: \(-\frac{1}{x} + c\)