int_1^15(x+2)/((x+1)(x+3))dx= কত?

Explanation:

Another Explanation (5): সমাধান: ধরি, \( \frac{x+2}{(x+1)(x+3)} = \frac{A}{x+1} + \frac{B}{x+3} \) \( \Rightarrow x+2 = A(x+3) + B(x+1) \) যদি \( x = -1 \) হয়, তবে \( -1+2 = A(-1+3) + B(-1+1) \) \( \Rightarrow 1 = 2A \) \( \Rightarrow A = \frac{1}{2} \) যদি \( x = -3 \) হয়, তবে \( -3+2 = A(-3+3) + B(-3+1) \) \( \Rightarrow -1 = -2B \) \( \Rightarrow B = \frac{1}{2} \) সুতরাং, \( \frac{x+2}{(x+1)(x+3)} = \frac{1/2}{x+1} + \frac{1/2}{x+3} = \frac{1}{2} \left( \frac{1}{x+1} + \frac{1}{x+3} \right) \) এখন, \( \int_1^{15} \frac{x+2}{(x+1)(x+3)} dx = \frac{1}{2} \int_1^{15} \left( \frac{1}{x+1} + \frac{1}{x+3} \right) dx \) \( = \frac{1}{2} \left[ \ln|x+1| + \ln|x+3| \right]_1^{15} \) \( = \frac{1}{2} \left[ \ln(x+1) + \ln(x+3) \right]_1^{15} \) \( = \frac{1}{2} \left[ \ln(15+1) + \ln(15+3) - \ln(1+1) - \ln(1+3) \right] \) \( = \frac{1}{2} \left[ \ln(16) + \ln(18) - \ln(2) - \ln(4) \right] \) \( = \frac{1}{2} \left[ \ln(16 \times 18) - \ln(2 \times 4) \right] \) \( = \frac{1}{2} \left[ \ln(288) - \ln(8) \right] \) \( = \frac{1}{2} \ln \left( \frac{288}{8} \right) \) \( = \frac{1}{2} \ln(36) \) \( = \frac{1}{2} \ln(6^2) \) \( = \frac{1}{2} \cdot 2 \ln(6) \) \( = \ln(6) \) সুতরাং, \( \int_1^{15} \frac{x+2}{(x+1)(x+3)} dx = \ln 6 \) 😃