If int x^3(log x)^2 d x = x^4[A(log x)^2 + B( | vedclass

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(D) We use the method of integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = (\log x)^2$ and $v = x^3$. Then $u' = 2

Vedclass · Accepted Answer

$\frac{5}{32}$

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(D) We use the method of integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = (\log x)^2$ and $v = x^3$. Then $u' = 2 \log x \cdot \frac{1}{x}$ and $\int v dx = \frac{x^4}{4}$. $\int x^3(\log x)^2 dx = \frac{x^4}{4}(\log x)^2 - \int (2 \log x \cdot \frac{1}{x} \cdot \frac{x^4}{4}) dx = \frac{x^4}{4}(\log x)^2 - \frac{1}{2} \int x^3 \log x dx$. Now, integrate $\int x^3 \log x dx$ by parts again: $\int x^3 \log x dx = \log x \cdot \frac{x^4}{4} - \int \frac{1}{x} \cdot \frac{x^4}{4} dx = \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 dx = \frac{x^4}{4} \log x - \frac{x^4}{16}$. Substituting this back: $\int x^3(\log x)^2 dx = \frac{x^4}{4}(\log x)^2 - \frac{1}{2} [\frac{x^4}{4} \log x - \frac{x^4}{16}] + K = x^4 [\frac{1}{4}(\log x)^2 - \frac{1}{8} \log x + \frac{1}{32}] + K$. Comparing with $x^4[A(\log x)^2 + B(\log x) + C]$, we get $A = \frac{1}{4}$, $B = -\frac{1}{8}$, and $C = \frac{1}{32}$. Thus, $A + B + C = \frac{1}{4} - \frac{1}{8} + \frac{1}{32} = \frac{8 - 4 + 1}{32} = \frac{5}{32}$.

If $\int x^3(\log x)^2 d x = x^4[A(\log x)^2 + B(\log x) + C] + K$, then find the value of $A + B + C$.

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