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$\int {(\log_{e} x)}^{3} d x =$

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x [{(\log_{e} x)}^{3} + {(\log_{e} x)}^{2} - (\log_{e} x) - 6] + c

x [{(\log_{e} x)}^{3} - {(\log_{e} x)}^{2} + (\log_{e} x) - 6] + c

x [{(\log_{e} x)}^{3} - 3 {(\log_{e} x)}^{2} + 6 (\log_{e} x) - 6] + c

x [{(\log_{e} x)}^{3} + {(\log_{e} x)}^{2} + (\log_{e} x) - 6] + c

answer is D.

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Detailed Solution

= \int {(\log_{e} x)}^{n} d x

In $= x {(\log_{e} x)}^{n} - n . I_{n - 1}$

Put $n = 3$

$I_{3} = x {(\log_{e} x)}^{3} - 3 I_{2}$

$= x {(\log_{e} x)}^{3} - 3 {x {(\log_{e} x)}^{2} - 2 I_{1}}$

$= x {(\log_{e} x)}^{3} - 3 x {(\log_{e} x)}^{2} + 6 I_{1}$

$= x {(\log_{e} x)}^{3} - 3 x {(\log_{e} x)}^{2} + 6 (x \log_{e} x - I_{0})$

$= x {(\log_{e} x)}^{3} - 3 x {(\log_{e} x)}^{2} + 6 x \log_{e} x - 6 x + c$

$= x {{(\log_{e} x)}^{3} - 3 {(\log_{e} x)}^{2} + 6 \log_{e} x - 6} + c$

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