int {{x^3}log x,dx = } | vedclass

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Question

(B) To solve the integral $I = \int {{x^3}\log x\,dx}$,we use the method of integration by parts: $\int {u\,dv} = uv - \int {v\,du}$.Let $u = \log x$

Vedclass · Accepted Answer

$\frac{1}{{16}}[4{x^4}\log x - {x^4}] + c$

Vedclass · Answer

(B) To solve the integral $I = \int {{x^3}\log x\,dx}$,we use the method of integration by parts: $\int {u\,dv} = uv - \int {v\,du}$.Let $u = \log x$ and $dv = x^3\,dx$.Then $du = \frac{1}{x}\,dx$ and $v = \frac{x^4}{4}$.Applying the formula:$I = \left( \log x \cdot \frac{x^4}{4} \right) - \int {\frac{x^4}{4} \cdot \frac{1}{x}\,dx} + c$$I = \frac{x^4}{4}\log x - \frac{1}{4}\int {x^3\,dx} + c$$I = \frac{x^4}{4}\log x - \frac{1}{4} \cdot \frac{x^4}{4} + c$$I = \frac{x^4}{4}\log x - \frac{x^4}{16} + c$Taking $\frac{1}{16}$ as a common factor:$I = \frac{1}{16}[4x^4\log x - x^4] + c$.

$\int {{x^3}\log x\,dx = } $

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