int frac{1}{x^3} [log x^x]^2 , dx = | vedclass

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Question

(B) Given integral: $I = \int \frac{1}{x^3} [\log x^x]^2 \, dx$ Since $\log x^x = x \log x$,we substitute this into the integral: $I = \int \frac{1}{x

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$\frac{1}{3}(\log x)^3 + c$

Vedclass · Answer

(B) Given integral: $I = \int \frac{1}{x^3} [\log x^x]^2 \, dx$ Since $\log x^x = x \log x$,we substitute this into the integral: $I = \int \frac{1}{x^3} (x \log x)^2 \, dx$ $I = \int \frac{1}{x^3} (x^2 (\log x)^2) \, dx$ $I = \int \frac{1}{x} (\log x)^2 \, dx$ Now,let $t = \log x$,then $dt = \frac{1}{x} \, dx$. Substituting these into the integral: $I = \int t^2 \, dt = \frac{t^3}{3} + c$ Replacing $t$ with $\log x$: $I = \frac{1}{3}(\log x)^3 + c$

$\int \frac{1}{x^3} [\log x^x]^2 \, dx = $

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