int (log x)^3 x^5 dx = | vedclass

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Question

(A) माना $I = \int (\log x)^3 x^5 dx$ है। खंडशः समाकलन (integration by parts) $\int u v dx = u \int v dx - \int (u' \int v dx) dx$ का उपयोग करते हुए,ज

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$x^6 \left[ \frac{(\log x)^3}{6} - \frac{(\log x)^2}{12} + \frac{\log x}{36} - \frac{1}{216} \right] + c$

Vedclass · Answer

(A) माना $I = \int (\log x)^3 x^5 dx$ है। खंडशः समाकलन (integration by parts) $\int u v dx = u \int v dx - \int (u' \int v dx) dx$ का उपयोग करते हुए,जहाँ $u = (\log x)^3$ और $v = x^5$ है। $I = (\log x)^3 \frac{x^6}{6} - \int 3(\log x)^2 \frac{1}{x} \frac{x^6}{6} dx = \frac{x^6}{6}(\log x)^3 - \frac{1}{2} \int x^5 (\log x)^2 dx$। $\int x^5 (\log x)^2 dx$ के लिए पुनः खंडशः समाकलन का उपयोग करने पर: $I = \frac{x^6}{6}(\log x)^3 - \frac{1}{2} \left[ (\log x)^2 \frac{x^6}{6} - \int 2(\log x) \frac{1}{x} \frac{x^6}{6} dx \right] = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{1}{6} \int x^5 \log x dx$। $\int x^5 \log x dx$ के लिए पुनः खंडशः समाकलन का उपयोग करने पर: $I = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{1}{6} \left[ \log x \frac{x^6}{6} - \int \frac{1}{x} \frac{x^6}{6} dx \right] = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{x^6}{36} \log x - \frac{1}{36} \int x^5 dx$। $I = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{x^6}{36} \log x - \frac{1}{36} \frac{x^6}{6} + c = x^6 \left[ \frac{(\log x)^3}{6} - \frac{(\log x)^2}{12} + \frac{\log x}{36} - \frac{1}{216} \right] + c$।

$\int (\log x)^3 x^5 dx = $

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