int_0^1(xdx)/(1+x^4)=A হলে A এর মান কোনটি?

ধরি, \(x^2 = t\)

সুতরাং, \(2xdx = dt \Rightarrow xdx = \frac{1}{2}dt\)

যখন \(x = 0\), তখন \(t = 0\)

যখন \(x = 1\), তখন \(t = 1\)

অতএব, \(A = \int_0^1 \frac{xdx}{1+x^4} = \int_0^1 \frac{\frac{1}{2}dt}{1+t^2} = \frac{1}{2}\int_0^1 \frac{dt}{1+t^2}\)

আমরা জানি, \(\int \frac{dx}{1+x^2} = \tan^{-1}(x) + C\)

সুতরাং, \(A = \frac{1}{2} [\tan^{-1}(t)]_0^1 = \frac{1}{2} [\tan^{-1}(1) - \tan^{-1}(0)]\)

আমরা জানি, \(\tan^{-1}(1) = \frac{\pi}{4}\) এবং \(\tan^{-1}(0) = 0\)

সুতরাং, \(A = \frac{1}{2} [\frac{\pi}{4} - 0] = \frac{\pi}{8}\) 🎉

সুতরাং, \(A = \frac{\pi}{8}\) 🥳