int_ -∞^0(e^xdx)/(1+e^(2x)) 

Explanation:

Another Explanation (5): সমাধান: \( \int_{-\infty}^{0} \frac{e^x}{1+e^{2x}} dx \) ধরি, \( u = e^x \). সুতরাং, \( du = e^x dx \). যখন \( x = -\infty \), \( u = e^{-\infty} = 0 \). যখন \( x = 0 \), \( u = e^0 = 1 \). তাহলে, সমাকলনটি হবে: \( \int_{0}^{1} \frac{1}{1+u^2} du \) আমরা জানি, \( \int \frac{1}{1+u^2} du = \arctan(u) + C \) সুতরাং, \( \int_{0}^{1} \frac{1}{1+u^2} du = \left[ \arctan(u) \right]_{0}^{1} \) \( = \arctan(1) - \arctan(0) \) \( = \frac{\pi}{4} - 0 \) \( = \frac{\pi}{4} \) অতএব, \( \int_{-\infty}^{0} \frac{e^x}{1+e^{2x}} dx = \frac{\pi}{4} \) 🎉🎉🎉