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Question

(C) Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = (\log x)^2$ and $dv = x^3 \, dx$. Then $du = 2(\log x) \cdot \frac{1}{x} \

Vedclass · Accepted Answer

$(\log x)^2 \frac{x^4}{4} - \frac{1}{2} \left[ (\log x) \frac{x^4}{4} - \frac{x^4}{16} \right] + C$

Vedclass · Answer

(C) Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = (\log x)^2$ and $dv = x^3 \, dx$. Then $du = 2(\log x) \cdot \frac{1}{x} \, dx$ and $v = \frac{x^4}{4}$. $I = \int x^3(\log x)^2 \, dx = (\log x)^2 \cdot \frac{x^4}{4} - \int \frac{x^4}{4} \cdot 2(\log x) \cdot \frac{1}{x} \, dx$ $I = \frac{x^4}{4}(\log x)^2 - \frac{1}{2} \int x^3 \log x \, dx$ Now,apply integration by parts again for $\int x^3 \log x \, dx$: Let $u = \log x$ and $dv = x^3 \, dx$. Then $du = \frac{1}{x} \, dx$ and $v = \frac{x^4}{4}$. $\int x^3 \log x \, dx = \frac{x^4}{4} \log x - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx = \frac{x^4}{4} \log x - \frac{x^4}{16}$. Substituting this back: $I = \frac{x^4}{4}(\log x)^2 - \frac{1}{2} \left[ \frac{x^4}{4} \log x - \frac{x^4}{16} \right] + C$.

$\int x^3(\log x)^2 \, dx = $

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