How do you evaluate the integral $\int \frac{e^{\frac{1}{x}}}{x^{2}}$ $int e^(1/x)/x^2$ ?

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Answer 1 · 2017-01-28T11:35:04

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

Substitute: $t = \frac{1}{x}$ $t=1/x$ , $d t = - \frac{1}{x^{2}} d x$ $dt = -1/x^2dx$

so we have:

$\int \frac{e^{\frac{1}{x}}}{x^{2}} d x = - \int e^{t} d t = - e^{t} + C$ $int e^(1/x)/x^2dx = -int e^tdt = -e^t + C$

and undoing the substitution:

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

How do you evaluate the integral ∫e1xx2int e^(1/x)/x^2?