int(x-3)/((1-2x)(1+x))dx এর মান কত ?

Explanation:

Another Explanation (5): সমাধান: ধরি, \[ \frac{x-3}{(1-2x)(1+x)} = \frac{A}{1-2x} + \frac{B}{1+x} \] উভয় পক্ষে \((1-2x)(1+x)\) দিয়ে গুণ করে পাই, \[ x-3 = A(1+x) + B(1-2x) \] এখন, \(x = -1\) বসালে পাই, \[ -1 - 3 = A(1-1) + B(1-2(-1)) \] \[ -4 = 3B \] \[ B = -\frac{4}{3} \] আবার, \(x = \frac{1}{2}\) বসালে পাই, \[ \frac{1}{2} - 3 = A(1+\frac{1}{2}) + B(1-2(\frac{1}{2})) \] \[ -\frac{5}{2} = \frac{3}{2}A \] \[ A = -\frac{5}{3} \] সুতরাং, \[ \frac{x-3}{(1-2x)(1+x)} = \frac{-\frac{5}{3}}{1-2x} + \frac{-\frac{4}{3}}{1+x} \] এখন, \[ \int \frac{x-3}{(1-2x)(1+x)} dx = \int \left( \frac{-\frac{5}{3}}{1-2x} + \frac{-\frac{4}{3}}{1+x} \right) dx \] \[ = -\frac{5}{3} \int \frac{1}{1-2x} dx - \frac{4}{3} \int \frac{1}{1+x} dx \] \[ = -\frac{5}{3} \cdot \frac{\ln|1-2x|}{-2} - \frac{4}{3} \ln|1+x| + C \] \[ = \frac{5}{6} \ln|1-2x| - \frac{4}{3} \ln|1+x| + C \] অতএব, \(\int \frac{x-3}{(1-2x)(1+x)} dx = \frac{5}{6} \ln|1-2x| - \frac{4}{3} \ln|1+x| + C\) 🥳🎉