text{If } int x^4(log x)^3 dx = x^5[A(log x)^ | vedclass

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(A) Using integration by parts $\int u v dx = u \int v dx - \int (u' \int v dx) dx$,where $u = (\log x)^3$ and $v = x^4$: $\int x^4(\log x)^3 dx = \fr

Vedclass · Accepted Answer

$\frac{2}{25}$

Vedclass · Answer

(A) Using integration by parts $\int u v dx = u \int v dx - \int (u' \int v dx) dx$,where $u = (\log x)^3$ and $v = x^4$: $\int x^4(\log x)^3 dx = \frac{x^5}{5}(\log x)^3 - \int 3(\log x)^2 \cdot \frac{1}{x} \cdot \frac{x^5}{5} dx = \frac{x^5}{5}(\log x)^3 - \frac{3}{5} \int x^4(\log x)^2 dx$ Applying parts again: $= \frac{x^5}{5}(\log x)^3 - \frac{3}{5} [\frac{x^5}{5}(\log x)^2 - \int 2(\log x) \cdot \frac{1}{x} \cdot \frac{x^5}{5} dx] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25}x^5(\log x)^2 + \frac{6}{25} \int x^4 \log x dx$ Applying parts again: $= \frac{x^5}{5}(\log x)^3 - \frac{3}{25}x^5(\log x)^2 + \frac{6}{25} [\frac{x^5}{5} \log x - \int \frac{1}{x} \cdot \frac{x^5}{5} dx] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25}x^5(\log x)^2 + \frac{6}{125}x^5 \log x - \frac{6}{125} \int x^4 dx$ $= x^5 [\frac{1}{5}(\log x)^3 - \frac{3}{25}(\log x)^2 + \frac{6}{125} \log x - \frac{6}{625}] + k$ Comparing coefficients: $A = \frac{1}{5}, B = -\frac{3}{25}, C = \frac{6}{125}, D = -\frac{6}{625}$ $A + B + C + 5D = \frac{1}{5} - \frac{3}{25} + \frac{6}{125} + 5(-\frac{6}{625}) = \frac{25 - 15 + 6 - 6}{125} = \frac{10}{125} = \frac{2}{25}$

$\text{If } \int x^4(\log x)^3 dx = x^5[A(\log x)^3 + B(\log x)^2 + C \log x + D] + k, \text{ then } A + B + C + 5D = $

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