If M=int_0^{infty} frac{log t}{1+t^3} d t and | vedclass

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(D) Given $M = \int_0^{\infty} \frac{\log t}{1+t^3} dt$ and $N = \int_{-\infty}^{\infty} \frac{t e^{2 t}}{1+e^{3 t}} dt$. For $M$,let $t = e^{-x}$,the

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$N=-M$

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(D) Given $M = \int_0^{\infty} \frac{\log t}{1+t^3} dt$ and $N = \int_{-\infty}^{\infty} \frac{t e^{2 t}}{1+e^{3 t}} dt$. For $M$,let $t = e^{-x}$,then $dt = -e^{-x} dx$. When $t = 0, x = \infty$ and when $t = \infty, x = -\infty$. Substituting these into $M$: $M = \int_{\infty}^{-\infty} \frac{\log(e^{-x})}{1+(e^{-x})^3} (-e^{-x}) dx = \int_{-\infty}^{\infty} \frac{-x}{1+e^{-3x}} e^{-x} dx$. Multiply numerator and denominator by $e^{3x}$: $M = \int_{-\infty}^{\infty} \frac{-x e^{2x}}{e^{3x} + 1} dx = -\int_{-\infty}^{\infty} \frac{x e^{2x}}{1+e^{3x}} dx$. Comparing this with $N$,we get $M = -N$,which implies $N = -M$.

If $M=\int_0^{\infty} \frac{\log t}{1+t^3} d t$ and $N=\int_{-\infty}^{\infty} \frac{t e^{2 t}}{1+e^{3 t}} d t$,then

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