For $x > 0$ , if $\int {(\log x)}^{5} d x = x [A {(\log x)}^{5} + B {(\log x)}^{4} + C {(\log x)}^{3} + D {(\log x)}^{2} + E (\log x) + F] +$ constant, then $A + B + C + D + E + F =$

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answer is A.

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Detailed Solution

$I_{n} = \int {(\log x)}^{n} d x = x {(\log x)}^{n} - n . I_{n - 1} n \geq 1$

$I_{5} =$ $\int {(\log x)}^{5} d x = x {(\log x)}^{5} - 5 I_{4}$

$= x {(\log x)}^{5} - 5 [{(x \log x)}^{4} - 4 I_{3}] = x {(\log x)}^{5} - 5 x {(\log x)}^{4} + 20 [x {(\log x)}^{3} 3 \frac{I}{2}] = x {(\log x)}^{5} - 5 x {(\log x)}^{4} + 20 x {(\log x)}^{3} - 60 (x {(\log x)}^{2} - 2 I) = x {(\log x)}^{5} - 5 x {(\log x)}^{4} + 20 x {(\log x)}^{3} - 60 {(x (\log x)}^{2} + 120 (x \log x - x) (I_{1} = \int \log x d x = x \log x - x) = x [{(\log x)}^{5} - 5 {(\log x)}^{4} + 20 {(\log x)}^{3} - 60 {(\log x)}^{2}] + 120 \log x - 120 \Rightarrow A = 1, B = - 5, C = 20, D = - 60, E = 120, F = - 120 ∴ A + B + C + D + E + F = - 44$