\( \int_{1}^{\alpha} \left( 2 + x\ln(x^2 + 5) \right)dx + \int_{1}^{\alpha} \left( 3 - x\ln(x^2 + 5) \right)dx = 30 \) α এর মান কত?

Explanation: Hints: \(\int_a^b (u + v) dx = \int_a^b u dx + \int_a^b v dx\) \(\int_a^b f(x) dx = F(b) - F(a)\) Solve: \(\int_1^\alpha \{2 + x\ln(x^2 + 5)\} dx + \int_1^\alpha \{3 - x\ln(x^2 + 5)\} dx = 30\) \[ \implies \int_1^\alpha 2 dx + \int_1^\alpha x \ln(x^2 + 5) dx + \int_1^\alpha 3 dx - \int_1^\alpha x \ln(x^2 + 5) dx = 30 \implies \int_1^\alpha 2 dx + \int_1^\alpha 3 dx = 30 \] \[ \implies [2x]_1^\alpha + [3x]_1^\alpha = 30 \implies [2\alpha - 2] + [3\alpha - 3] = 30 \implies 5\alpha - 5 = 30 \implies 5\alpha = 35 \implies \alpha = 7 \] Ans. (E)

Another Explanation (5): bài giải: দেওয়া আছে, \[ \int_{1}^{\alpha} \left( 2 + x\ln(x^2 + 5) \right)dx + \int_{1}^{\alpha} \left( 3 - x\ln(x^2 + 5) \right)dx = 30 \] এখন, ইন্টিগ্রালগুলোকে একত্রিত করে পাই, \[ \int_{1}^{\alpha} \left[ (2 + x\ln(x^2 + 5)) + (3 - x\ln(x^2 + 5)) \right]dx = 30 \] \[ \int_{1}^{\alpha} (2 + x\ln(x^2 + 5) + 3 - x\ln(x^2 + 5)) dx = 30 \] \[ \int_{1}^{\alpha} (5) dx = 30 \] \[ 5 \int_{1}^{\alpha} dx = 30 \] \[ 5 [x]_{1}^{\alpha} = 30 \] \[ 5 (\alpha - 1) = 30 \] \[ \alpha - 1 = \frac{30}{5} \] \[ \alpha - 1 = 6 \] \[ \alpha = 6 + 1 \] \[ \alpha = 7 \] সুতরাং, \( \alpha \) এর মান 7। 🎉🎉🎉