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

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Question

(A) Let $I = \int (\log x)^3 x^4 \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = (\log x)^3$ and $dv = x^4 \, dx$. The

Vedclass · Accepted Answer

$\frac{x^5}{625} [125 p^3 - 75 p^2 + 30 p - 6] + c$ (where,$p = \log x$)

Vedclass · Answer

(A) Let $I = \int (\log x)^3 x^4 \, dx$. Using 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 $dv = x^4 \, dx$:$I = \frac{x^5}{5}(\log x)^3 - \frac{3}{5} [\frac{x^5}{5}(\log x)^2 - \int \frac{x^5}{5} \cdot 2(\log x) \cdot \frac{1}{x} \, dx] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25} x^5(\log x)^2 + \frac{6}{25} \int x^4 \log x \, dx$.Applying integration by parts for $\int x^4 \log x \, dx$ with $u = \log x$ and $dv = x^4 \, dx$:$I = \frac{x^5}{5}(\log x)^3 - \frac{3}{25} x^5(\log x)^2 + \frac{6}{25} [\frac{x^5}{5} \log x - \int \frac{x^5}{5} \cdot \frac{1}{x} \, dx] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25} x^5(\log x)^2 + \frac{6}{125} x^5 \log x - \frac{6}{625} x^5 + c$.Factoring out $\frac{x^5}{625}$:$I = \frac{x^5}{625} [125(\log x)^3 - 75(\log x)^2 + 30(\log x) - 6] + c$.Substituting $p = \log x$,we get $\frac{x^5}{625} [125 p^3 - 75 p^2 + 30 p - 6] + c$.

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

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