∫ 1 2 e 1 x x 2 d x {\displaystyle \int _{1}^{2}{\frac {e^{\frac {1}{x}}}{x^{2}}}\,dx}
u = 1 x d u = − 1 x 2 d x − d u = 1 x 2 d x {\displaystyle {\begin{aligned}u&={\frac {1}{x}}\\[2ex]du&=-{\frac {1}{x^{2}}}dx\\[2ex]-du&={\frac {1}{x^{2}}}dx\\[2ex]\end{aligned}}}
New upper limit: 1 2 = 1 ( 2 ) {\displaystyle {\frac {1}{2}}={\frac {1}{(2)}}} New lower limit: 1 = 1 ( 1 ) {\displaystyle 1={\frac {1}{(1)}}}
∫ 1 2 e 1 x x 2 d x = ∫ 1 2 e 1 x ( 1 x 2 d x ) = ∫ 1 1 2 e u ( − d u ) = − ∫ 1 1 2 e u d u = − e u | 1 1 2 = − e − ( − e 1 ) = e − e {\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {e^{\frac {1}{x}}}{x^{2}}}\,dx&=\int _{1}^{2}e^{\frac {1}{x}}({\frac {1}{x^{2}}}\,dx)&=\int _{1}^{\frac {1}{2}}e^{u}\,(-du)\\[2ex]&=-\int _{1}^{\frac {1}{2}}e^{u}\,du\\[2ex]&=-e^{u}{\bigg |}_{1}^{\frac {1}{2}}\\[2ex]&=-{\sqrt {e}}-(-e^{1})\\[2ex]&=e-{\sqrt {e}}\end{aligned}}}
![{\displaystyle {\begin{aligned}u&={\frac {1}{x}}\\[2ex]du&=-{\frac {1}{x^{2}}}dx\\[2ex]-du&={\frac {1}{x^{2}}}dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a2af755923cf83f08ed479cf396012097de2ef31)
![{\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {e^{\frac {1}{x}}}{x^{2}}}\,dx&=\int _{1}^{2}e^{\frac {1}{x}}({\frac {1}{x^{2}}}\,dx)&=\int _{1}^{\frac {1}{2}}e^{u}\,(-du)\\[2ex]&=-\int _{1}^{\frac {1}{2}}e^{u}\,du\\[2ex]&=-e^{u}{\bigg |}_{1}^{\frac {1}{2}}\\[2ex]&=-{\sqrt {e}}-(-e^{1})\\[2ex]&=e-{\sqrt {e}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/02709f129e72cb8b45327f6953de5371d0707195)