For x 0,if int (log x)^5 dx is equal to x[A | vedclass

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(A) Let $I = \int (\log x)^5 dx$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then $I = \int t^5 e^t dt$. Using integration by

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$-44$

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(A) Let $I = \int (\log x)^5 dx$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then $I = \int t^5 e^t dt$. Using integration by parts repeatedly,$\int t^n e^t dt = e^t [t^n - n t^{n-1} + n(n-1) t^{n-2} - \dots + (-1)^n n!]$. For $n = 5$,we have $I = e^t [t^5 - 5t^4 + 20t^3 - 60t^2 + 120t - 120] + C$. Substituting $t = \log x$ back,we get $I = x [(\log x)^5 - 5(\log x)^4 + 20(\log x)^3 - 60(\log x)^2 + 120(\log x) - 120] + C$. Comparing this with the given expression,we identify the coefficients: $A = 1, B = -5, C = 20, D = -60, E = 120, F = -120$. Thus,$A + B + C + D + E + F = 1 - 5 + 20 - 60 + 120 - 120 = -44$.

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

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