int (log_{e} 2x)^3 dx = | vedclass

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(A) Let $I = \int (\log_{e} 2x)^3 dx$. Substitute $t = \log_{e} 2x$,then $2x = e^t$,so $x = \frac{1}{2} e^t$ and $dx = \frac{1}{2} e^t dt$. $I = \int

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

Vedclass · Answer

(A) Let $I = \int (\log_{e} 2x)^3 dx$. Substitute $t = \log_{e} 2x$,then $2x = e^t$,so $x = \frac{1}{2} e^t$ and $dx = \frac{1}{2} e^t dt$. $I = \int t^3 \cdot \frac{1}{2} e^t dt = \frac{1}{2} \int t^3 e^t dt$. Using integration by parts $\int u dv = uv - \int v du$,where $u = t^3$ and $dv = e^t dt$: $I = \frac{1}{2} [t^3 e^t - \int 3t^2 e^t dt] = \frac{1}{2} t^3 e^t - \frac{3}{2} \int t^2 e^t dt$. Applying integration by parts again: $\int t^2 e^t dt = t^2 e^t - 2 \int t e^t dt = t^2 e^t - 2(t e^t - e^t) = t^2 e^t - 2t e^t + 2e^t$. Substituting back: $I = \frac{1}{2} t^3 e^t - \frac{3}{2} [t^2 e^t - 2t e^t + 2e^t] + c = \frac{1}{2} e^t [t^3 - 3t^2 + 6t - 6] + c$. Since $e^t = 2x$ and $t = \log_{e} 2x$: $I = \frac{1}{2} (2x) [(\log_{e} 2x)^3 - 3(\log_{e} 2x)^2 + 6(\log_{e} 2x) - 6] + c = x[(\log_{e} 2x)^3 - 3(\log_{e} 2x)^2 + 6(\log_{e} 2x) - 6] + c$.

$\int (\log_{e} 2x)^3 dx =$

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