If int(log x)^3 x^5 d x=frac{x^6}{A}left[B(lo | vedclass

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(C) We use integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = (\log x)^3$ and $v = x^5$. $\int (\log x)^3 x^5 dx = (

Vedclass · Accepted Answer

$192$

Vedclass · Answer

(C) We use integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = (\log x)^3$ and $v = x^5$. $\int (\log x)^3 x^5 dx = (\log x)^3 \frac{x^6}{6} - \int 3(\log x)^2 \frac{1}{x} \frac{x^6}{6} dx = \frac{x^6}{6}(\log x)^3 - \frac{1}{2} \int x^5 (\log x)^2 dx$. Applying integration by parts again: $= \frac{x^6}{6}(\log x)^3 - \frac{1}{2} [(\log x)^2 \frac{x^6}{6} - \int 2(\log x) \frac{1}{x} \frac{x^6}{6} dx] = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{1}{6} \int x^5 \log x dx$. Applying integration by parts once more: $= \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{1}{6} [(\log x) \frac{x^6}{6} - \int \frac{1}{x} \frac{x^6}{6} dx] = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{x^6}{36}\log x - \frac{1}{36} \int x^5 dx$. $= \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{x^6}{36}\log x - \frac{x^6}{216} + k$. Factor out $\frac{x^6}{216}$: $= \frac{x^6}{216} [36(\log x)^3 - 18(\log x)^2 + 6(\log x) - 1] + k$. Comparing with the given form, $A = 216, B = 36, C = -18, D = 6$. Thus, $A - (B + C + D) = 216 - (36 - 18 + 6) = 216 - 24 = 192$.

If $\int(\log x)^3 x^5 d x=\frac{x^6}{A}\left[B(\log x)^3+C(\log x)^2+D(\log x)-1\right]+k$ and $A, B, C, D$ are integers, then $A-(B+C+D)=$

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