int_(e^2)^e (lnx)^2 dx =?  

Explanation:

Another Explanation (5): সমাধান: \( \int_{e^2}^{e} (\ln x)^2 dx \) নির্ণয় করতে হবে। ধরি, \(I = \int_{e^2}^{e} (\ln x)^2 dx \) আমরা parts এর মাধ্যমে ইন্টিগ্রেশন করব: \( \int u v dx = u \int v dx - \int ( \frac{du}{dx} \int v dx ) dx \) এখানে, \( u = (\ln x)^2 \) এবং \( v = 1 \) তাহলে, \( I = (\ln x)^2 \int_{e^2}^{e} 1 dx - \int_{e^2}^{e} \frac{d}{dx} (\ln x)^2 \int 1 dx \; dx \) \( = [(\ln x)^2 x]_{e^2}^{e} - \int_{e^2}^{e} 2 \ln x \cdot \frac{1}{x} \cdot x \; dx \) \( = [(\ln e)^2 e - (\ln e^2)^2 e^2] - 2 \int_{e^2}^{e} \ln x \; dx \) \( = [1^2 \cdot e - (2)^2 e^2] - 2 \int_{e^2}^{e} \ln x \; dx \) \( = [e - 4e^2] - 2 \int_{e^2}^{e} \ln x \; dx \) এখন, \( \int_{e^2}^{e} \ln x \; dx \) নির্ণয় করি। আবার parts এর মাধ্যমে ইন্টিগ্রেশন করে পাই: \( \int \ln x \cdot 1 \; dx = \ln x \int 1 dx - \int \frac{d}{dx} (\ln x) \int 1 dx \; dx \) \( = x \ln x - \int \frac{1}{x} \cdot x \; dx \) \( = x \ln x - \int 1 \; dx \) \( = x \ln x - x + C \) সুতরাং, \( \int_{e^2}^{e} \ln x \; dx = [x \ln x - x]_{e^2}^{e} \) \( = [e \ln e - e] - [e^2 \ln e^2 - e^2] \) \( = [e \cdot 1 - e] - [e^2 \cdot 2 - e^2] \) \( = [e - e] - [2e^2 - e^2] \) \( = 0 - e^2 = -e^2 \) অতএব, \( I = e - 4e^2 - 2(-e^2) \) \( = e - 4e^2 + 2e^2 \) \( = e - 2e^2 \) সুতরাং, \( \int_{e^2}^{e} (\ln x)^2 dx = e - 2e^2 \) final answer is \(e - 2e^2\)🎉.