Word Count : 401

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हमारे पास परिभाषित इंटीग्रल है: [ I_{m,n} = \int x^m (\log x)^n , dx ] इसे साबित करने के लिए कि [ (m+1) I_{m,n} = x^{m+1} (\log x)^n - n I_{m,n-1}, ] हम इंटीग्रेशन बाय पार्ट्स (Integration by Parts) का प्रयोग करते हैं। इंटीग्रेशन बाय पार्ट्स का सूत्र है: [ \int u , dv = uv - \int v , du ] हम चुनते हैं: [ u = (\log x)^n \quad \Rightarrow \quad du = n (\log x)^{n-1} \frac{dx}{x}, ] [ dv = x^m dx \quad \Rightarrow \quad v = \frac{x^{m+1}}{m+1} \quad (m \neq -1). ] अब बाय पार्ट्स लागू _________ __________ _________ _______ ______ __________ _________ ____ ________ _____ ________.
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