How do you integrate $\int \frac{e^{\frac{1}{x}}}{x^{2}}$ $inte^(1/x)/x^2$ using substitution?

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Answer 1 · 2017-02-16T08:24:36

$- e^{\frac{1}{x}} + C$ $-e^(1/x)+C$

$I = \int \frac{e^{\frac{1}{x}}}{x^{2}} d x$ $I = int e^(1/x)/x^2 dx$

Let $u = \frac{1}{x} \to \frac{d u}{d x} = - \frac{1}{x^{2}}$ $u=1/x -> (du)/dx=-1/x^2$

$:. dx=-x^2du$

$I = -int e^u/x^2 * x^2 du$

$= -int e^u/ cancel (x^2) * cancel(x^2) du$

$=-e^u + C = -e^(1/x)+C$

Answer 2 · 2017-02-16T08:24:52

$int e^(1/x)/x^2dx= -e^(1/x)+C$

Substitute $t=1/x$ , $dt=-(dx)/x^2$

$int e^(1/x)/x^2dx= -int e^tdt = -e^t+C$

and undoing the substitution:

$int e^(1/x)/x^2dx= -e^(1/x)+C$

How do you integrate inte^(1/x)/x^2∫e1xx2inte^(1/x)/x^2 using substitution?