int(log x)^2 x^3 d x=frac{x^4}{32} f(x)+C Rig | vedclass

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Question

(A) Let $I = \int (\log x)^2 x^3 dx$.Using 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

Vedclass · Accepted Answer

$8(\log x)^2-4 \log x+1$

Vedclass · Answer

(A) Let $I = \int (\log x)^2 x^3 dx$.Using 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 $u' = \frac{2 \log x}{x}$ and $\int v dx = \frac{x^4}{4}$.$I = (\log x)^2 \frac{x^4}{4} - \int \frac{2 \log x}{x} \cdot \frac{x^4}{4} dx$$I = \frac{x^4}{4} (\log x)^2 - \frac{1}{2} \int x^3 \log x dx$.Now,evaluate $\int x^3 \log x dx$ using integration by parts again with $u = \log x$ and $v = x^3$.$\int x^3 \log x dx = (\log x) \frac{x^4}{4} - \int \frac{1}{x} \cdot \frac{x^4}{4} dx$$= \frac{x^4 \log x}{4} - \frac{1}{4} \int x^3 dx = \frac{x^4 \log x}{4} - \frac{x^4}{16}$.Substituting this back into the expression for $I$:$I = \frac{x^4}{4} (\log x)^2 - \frac{1}{2} \left( \frac{x^4 \log x}{4} - \frac{x^4}{16} \right) + C$$I = \frac{x^4}{4} (\log x)^2 - \frac{x^4 \log x}{8} + \frac{x^4}{32} + C$Factor out $\frac{x^4}{32}$:$I = \frac{x^4}{32} \left( 8(\log x)^2 - 4 \log x + 1 \right) + C$.Comparing this with $\frac{x^4}{32} f(x) + C$,we get $f(x) = 8(\log x)^2 - 4 \log x + 1$.

$\int(\log x)^2 x^3 d x=\frac{x^4}{32} f(x)+C \Rightarrow f(x)=$

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