int (log x)^3 x^4 , dx = | vedclass

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(A) To evaluate $I = \int (\log x)^3 x^4 \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$.Let $u = (\log x)^3$ and $dv = x^4 \,

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$x^5 \left[ \frac{1}{5}(\log x)^3 - \frac{3}{25}(\log x)^2 + \frac{6}{125} \log x - \frac{6}{625} \right] + c$

Vedclass · Answer

(A) To evaluate $I = \int (\log x)^3 x^4 \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$.Let $u = (\log x)^3$ and $dv = x^4 \, dx$.Then $du = 3(\log x)^2 \cdot \frac{1}{x} \, dx$ and $v = \frac{x^5}{5}$.$I = \frac{x^5}{5}(\log x)^3 - \int \frac{x^5}{5} \cdot 3(\log x)^2 \cdot \frac{1}{x} \, dx = \frac{x^5}{5}(\log x)^3 - \frac{3}{5} \int x^4 (\log x)^2 \, dx$.Applying integration by parts again for $\int x^4 (\log x)^2 \, dx$ with $u = (\log x)^2$ and $dv = x^4 \, dx$:$I = \frac{x^5}{5}(\log x)^3 - \frac{3}{5} \left[ \frac{x^5}{5}(\log x)^2 - \int \frac{x^5}{5} \cdot 2 \log x \cdot \frac{1}{x} \, dx \right] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25} x^5 (\log x)^2 + \frac{6}{25} \int x^4 \log x \, dx$.Applying integration by parts one last time for $\int x^4 \log x \, dx$ with $u = \log x$ and $dv = x^4 \, dx$:$I = \frac{x^5}{5}(\log x)^3 - \frac{3}{25} x^5 (\log x)^2 + \frac{6}{25} \left[ \frac{x^5}{5} \log x - \int \frac{x^5}{5} \cdot \frac{1}{x} \, dx \right] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25} x^5 (\log x)^2 + \frac{6}{125} x^5 \log x - \frac{6}{125} \int x^4 \, dx$.$I = \frac{x^5}{5}(\log x)^3 - \frac{3}{25} x^5 (\log x)^2 + \frac{6}{125} x^5 \log x - \frac{6}{625} x^5 + c$.Factoring out $x^5$,we get $x^5 \left[ \frac{1}{5}(\log x)^3 - \frac{3}{25}(\log x)^2 + \frac{6}{125} \log x - \frac{6}{625} \right] + c$.

$\int (\log x)^3 x^4 \, dx =$

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