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(B) Let $I = \int (\log x)^3 dx$. Substitute $\log x = t$, which implies $x = e^t$ and $dx = e^t dt$. Then, $I = \int t^3 e^t dt$. Using the integrati

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$x [(\log x)^3 - 3(\log x)^2 + 6 \log x - 6] + c$

Vedclass · Answer

(B) Let $I = \int (\log x)^3 dx$. Substitute $\log x = t$, which implies $x = e^t$ and $dx = e^t dt$. Then, $I = \int t^3 e^t dt$. Using the integration by parts formula $\int e^t f(t) dt = e^t [f(t) - f'(t) + f''(t) - f'''(t) + \dots]$, where $f(t) = t^3$: $f'(t) = 3t^2$, $f''(t) = 6t$, and $f'''(t) = 6$. Therefore, $I = e^t [t^3 - 3t^2 + 6t - 6] + C$. Substituting back $t = \log x$ and $e^t = x$: $I = x [(\log x)^3 - 3(\log x)^2 + 6 \log x - 6] + C$.

$\int (\log x)^3 dx = $

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