How do you integrate e^(7/x)/x^2 ?

Answer 1

- e(7/x)/7 + c

Integrate by parts

Answer 2

e^(7/x)/x^2
put 1/x=k
-dx/x^2=dk
dx=-x^2dk
so
e^(7k) *-x^2dk/x^2
so we get
-e^7k dk
so by integrating we get
-e^(7k) / 7
substitute k=1/x
so final answer is -e^(7/x) / 7

Answer 3

int. [e^(7/x)] / x^2 dx
use u= 1/x
then du/dx = -1/x^2
the integral becomes,
int. -e^(7u) du = -[e^7u]/7 + c

Answer 4

∫[e^(7/x)] / x² dx =

u = 1/x
du = -1/x² dx

=>
∫[e^(7/x)] / x² dx = -∫[e^(7u)] du =

t = 7u
dt = 7 du
dt/7 = du

=>
∫[e^(7/x)] / x² dx = -∫[e^(7u)] du = -1/7∫e^t dt = -1/7e^t + c
∫[e^(7/x)] / x² dx = -∫[e^(7u)] du = -1/7e^(7u) + c

∫[e^(7/x)] / x² dx = -(1/7)e^(7/x) + c

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