int_0^1 x/(1+x^4) dx এর মান কত? 

ধরি, \(I = \int_0^1 \frac{x}{1+x^4} dx\).

এখন, \(x^2 = t\) ধরলে, \(2x dx = dt\) হয়। সুতরাং, \(x dx = \frac{1}{2} dt\).

যখন \(x = 0\), তখন \(t = 0^2 = 0\). এবং যখন \(x = 1\), তখন \(t = 1^2 = 1\).

অতএব, \(I = \int_0^1 \frac{1}{1+t^2} \cdot \frac{1}{2} dt = \frac{1}{2} \int_0^1 \frac{1}{1+t^2} dt\).

আমরা জানি, \(\int \frac{1}{1+t^2} dt = \tan^{-1}(t) + C\).

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

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

সুতরাং, \(I = \frac{1}{2} \left[ \frac{\pi}{4} - 0 \right] = \frac{\pi}{8}\).

অতএব, \(\int_0^1 \frac{x}{1+x^4} dx = \frac{\pi}{8}\). 🎉