Evaluate the integral: int x^3 log x , dx | vedclass

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(A) Using 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{

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Using 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 $v = \frac{x^4}{4}$.$\int x^3 \log x \, dx = (\log x) \left( \frac{x^4}{4} \right) - \int \left( \frac{x^4}{4} \right) \left( \frac{1}{x} \right) \, dx$.$= \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 \, dx$.$= \frac{x^4}{4} \log x - \frac{1}{4} \left( \frac{x^4}{4} \right) + c$.$= \frac{x^4}{4} \log x - \frac{x^4}{16} + c$.$= \frac{x^4}{16} [4 \log x - 1] + c$.

Evaluate the integral: $\int x^3 \log x \, dx$

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