How do you find the indefinite integral of $int ((6e^(1/x))/x^2)$ ?

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Answer 1 · 2015-10-19T16:50:40

$int(6e^(1/x))/x^2dx=-6e^(1/x)+C$

You can calculate this integral by substitution $t=1/x$

$int(6e^(1/x))/x^2dx=|(t=1/x),(dt=-1/x^2dx),(-dt=dx/x^2)|=$

$=int6e^t*(-dt)=-6inte^tdt=-6e^t+C=-6e^(1/x)+C$

How do you find the indefinite integral of int ((6e^(1/x))/x^2)int ((6e^(1/x))/x^2)?