[tex]\bf \displaystyle \int\limits_{1}^{e}\cfrac{1}{t}\cdot dt\\\\
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\textit{doing substitution}\\\\
u=\cfrac{1}{t}\implies u=t^{-1}\implies \cfrac{du}{dt}=-t^{-2}\implies \cfrac{du}{dt}=-\cfrac{1}{t^2}\\\\\\ -t^2du=dt\\\\
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\displaystyle \int\limits_{1}^{e}u\cdot -t^2du\impliedby \textit{now, let's do some substitution on the "t"}\\\\
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[tex]\bf u=\cfrac{1}{t}\implies t=\cfrac{1}{u}\implies t^2=\cfrac{1^2}{u^2}\implies t^2=\cfrac{1}{u^2}\\\\
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\displaystyle \int\limits_{1}^{e}u\cdot -\cfrac{1}{u^2}\cdot du\implies -1\int\limits_{1}^{e}\cfrac{1}{u}\cdot du\implies \left. -ln|u| \cfrac{}{}\right]_1^e[/tex]
[tex]\bf \left. -ln\left( \frac{1}{t} \right) \cfrac{}{}\right]_1^e\implies
\left[ -ln\left( \frac{1}{e} \right) \right]-\left[ -ln\left( \frac{1}{1} \right) \right]\implies
\left[ -ln\left( e^{-1}\right) \right]-\left[ -ln\left( 1\right) \right]
\\\\\\\
[-(-1)]-[-(0)]\implies 1-0\implies 1[/tex]
