int_0^1dx/(sqrt(x+1)+sqrtx) এর মান কত? 

Explanation:

Another Explanation (5): সমাধান: আমরা প্রথমে ইন্টিগ্রালটিকে সরল করি: \[ \int_0^1 \frac{dx}{\sqrt{x+1} + \sqrt{x}} \] হর থেকে রুট দূর করার জন্য, আমরা \(\sqrt{x+1} - \sqrt{x}\) দিয়ে লব ও হরকে গুণ করি: \[ \int_0^1 \frac{\sqrt{x+1} - \sqrt{x}}{(\sqrt{x+1} + \sqrt{x})(\sqrt{x+1} - \sqrt{x})} dx \] \[ \int_0^1 \frac{\sqrt{x+1} - \sqrt{x}}{(x+1) - x} dx \] \[ \int_0^1 (\sqrt{x+1} - \sqrt{x}) dx \] এখন আমরা ইন্টিগ্রেশন করি: \[ \int_0^1 (\sqrt{x+1} - \sqrt{x}) dx = \int_0^1 (x+1)^{1/2} dx - \int_0^1 x^{1/2} dx \] \[ = \left[ \frac{(x+1)^{3/2}}{3/2} \right]_0^1 - \left[ \frac{x^{3/2}}{3/2} \right]_0^1 \] \[ = \frac{2}{3} \left[ (x+1)^{3/2} \right]_0^1 - \frac{2}{3} \left[ x^{3/2} \right]_0^1 \] এখন আমরা লিমিট বসাই: \[ = \frac{2}{3} \left[ (1+1)^{3/2} - (0+1)^{3/2} \right] - \frac{2}{3} \left[ 1^{3/2} - 0^{3/2} \right] \] \[ = \frac{2}{3} \left[ 2^{3/2} - 1^{3/2} \right] - \frac{2}{3} \left[ 1 - 0 \right] \] \[ = \frac{2}{3} \left[ 2\sqrt{2} - 1 \right] - \frac{2}{3} \] \[ = \frac{2}{3} (2\sqrt{2} - 1) - \frac{2}{3} \] \[ = \frac{4\sqrt{2}}{3} - \frac{2}{3} - \frac{2}{3} \] \[ = \frac{4\sqrt{2}}{3} - \frac{4}{3} \] \[ = \frac{4\sqrt{2} - 4}{3} \] \[ = \frac{4}{3} (\sqrt{2} - 1) = \frac{2}{3} (2\sqrt{2} - 2) \] সুতরাং, \[ \int_0^1 \frac{dx}{\sqrt{x+1} + \sqrt{x}} = \frac{2}{3} (2\sqrt{2} - 2) \] অতএব, উত্তর: \(\frac{2}{3} (2\sqrt{2} - 2)\) 🎉