$\displaystyle \int \frac{e^\frac{1}{t}}{t^2}dt$ u=$\displaystyle e^{\frac{1}{t}}$ dv=$\displaystyle t^{-2}$ Is the substitution right on this one? Sorry for asking some many of the same types of questions...
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Originally Posted by saiyanmx89 $\displaystyle \int \frac{e^\frac{1}{t}}{t^2}dt$ u=$\displaystyle e^{\frac{1}{t}}$ dv=$\displaystyle t^{-2}$ Is the substitution right on this one? Sorry for asking some many of the same types of questions... It's not an integration by parts. if you make your substitution $\displaystyle u = e^{\frac{1}{t}},\;\;\text{then}\;\; du = - \frac{1}{t^2} e^{\frac{1}{t}}\,dt$ and your integral becomes really simple.
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