int_0^{infty} (x^{12} + x^{-12}) frac{log x}{ | vedclass

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(A) Let $I = \int_0^{\infty} (x^{12} + x^{-12}) \frac{\log x}{x} dx$. Substitute $x = \frac{1}{t}$,then $dx = -\frac{1}{t^2} dt$. As $x 	o 0$,$t 	o

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(A) Let $I = \int_0^{\infty} (x^{12} + x^{-12}) \frac{\log x}{x} dx$. Substitute $x = \frac{1}{t}$,then $dx = -\frac{1}{t^2} dt$. As $x 	o 0$,$t 	o \infty$,and as $x 	o \infty$,$t 	o 0$. Also,$\log x = \log(t^{-1}) = -\log t$. Substituting these into the integral: $I = \int_{\infty}^0 (t^{-12} + t^{12}) \frac{-\log t}{1/t} \left(-\frac{1}{t^2}ight) dt$. $I = \int_{\infty}^0 (t^{-12} + t^{12}) \frac{-\log t}{t} dt$. $I = -\int_0^{\infty} (t^{-12} + t^{12}) \frac{\log t}{t} dt$. Since the variable of integration is a dummy variable,we can replace $t$ with $x$: $I = -\int_0^{\infty} (x^{-12} + x^{12}) \frac{\log x}{x} dx$. $I = -I$. $2I = 0 \Rightarrow I = 0$.

$\int_0^{\infty} (x^{12} + x^{-12}) \frac{\log x}{x} dx =$

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