∫1 2dx/(x-1)^(2⁄3) la integral impropia los límites son de 1 hasta 2 de dx/(x-1) elevado a 2/3
CarlosMath
![\displaystyle I=\int\limits_{1}^{2}\dfrac{1}{(x-1)^{2/3}}dx\\ \\ \\ I=\lim\limits_{a\to1^+}\int\limits_{a}^{2}\dfrac{1}{(x-1)^{2/3}}dx\\ \\ \\ I=\lim\limits_{a\to1^+} \left.\left[3(x-1)^{1/3}\right]\right|_{a}^{2}\\\\\\ I=3\lim\limits_{a\to1^+} 1-(a-1)^{1/3}\\\\\\ \boxed{I=3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%0AI%3D%5Cint%5Climits_%7B1%7D%5E%7B2%7D%5Cdfrac%7B1%7D%7B%28x-1%29%5E%7B2%2F3%7D%7Ddx%5C%5C+%5C%5C+%5C%5C%0AI%3D%5Clim%5Climits_%7Ba%5Cto1%5E%2B%7D%5Cint%5Climits_%7Ba%7D%5E%7B2%7D%5Cdfrac%7B1%7D%7B%28x-1%29%5E%7B2%2F3%7D%7Ddx%5C%5C+%5C%5C+%5C%5C%0AI%3D%5Clim%5Climits_%7Ba%5Cto1%5E%2B%7D+%5Cleft.%5Cleft%5B3%28x-1%29%5E%7B1%2F3%7D%5Cright%5D%5Cright%7C_%7Ba%7D%5E%7B2%7D%5C%5C%5C%5C%5C%5C%0AI%3D3%5Clim%5Climits_%7Ba%5Cto1%5E%2B%7D+1-%28a-1%29%5E%7B1%2F3%7D%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BI%3D3%7D "\displaystyle I=\int\limits_{1}^{2}\dfrac{1}{(x-1)^{2/3}}dx\\ \\ \\ I=\lim\limits_{a\to1^+}\int\limits_{a}^{2}\dfrac{1}{(x-1)^{2/3}}dx\\ \\ \\ I=\lim\limits_{a\to1^+} \left.\left[3(x-1)^{1/3}\right]\right|_{a}^{2}\\\\\\ I=3\lim\limits_{a\to1^+} 1-(a-1)^{1/3}\\\\\\ \boxed{I=3}")
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wilpec
la expresión es dx/(x-1)^2/3
CarlosMath
corregido
wilpec
Gracias si esa es la respuesta
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