int {32{x^3}{{(log x)}^2}dx} is equal to | vedclass

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(A) Let $I = \int {32{x^3}{{(\log x)}^2}} dx = 32\int {{x^3}{{(\log x)}^2}dx} $.Using integration by parts,$\int u v dx = u \int v dx - \int (u' \int

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

Vedclass · Answer

(A) Let $I = \int {32{x^3}{{(\log x)}^2}} dx = 32\int {{x^3}{{(\log x)}^2}dx} $.Using integration by parts,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$,where $u = {(\log x)^2}$ and $v = {x^3}$.$I = 32 \left[ {(\log x)^2 \cdot \frac{x^4}{4} - \int {2 \log x \cdot \frac{1}{x} \cdot \frac{x^4}{4} dx} } \right]$$I = 32 \left[ \frac{x^4 (\log x)^2}{4} - \frac{1}{2} \int x^3 \log x dx \right]$Now,apply integration by parts again for $\int x^3 \log x dx$ with $u = \log x$ and $v = x^3$:$I = 32 \left[ \frac{x^4 (\log x)^2}{4} - \frac{1}{2} \left( \log x \cdot \frac{x^4}{4} - \int \frac{1}{x} \cdot \frac{x^4}{4} dx \right) \right] + c$$I = 32 \left[ \frac{x^4 (\log x)^2}{4} - \frac{x^4 \log x}{8} + \frac{1}{8} \int x^3 dx \right] + c$$I = 32 \left[ \frac{x^4 (\log x)^2}{4} - \frac{x^4 \log x}{8} + \frac{x^4}{32} \right] + c$$I = 8x^4 (\log x)^2 - 4x^4 \log x + x^4 + c$$I = x^4 \{ 8(\log x)^2 - 4 \log x + 1 \} + c$.

$\int {32{x^3}{{(\log x)}^2}dx} $ is equal to

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