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

Answer 1

Paisley Johnson

#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#

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Answer 2

Emilia Aguilar

To find the indefinite integral of ( \int \frac{6e^{1/x}}{x^2} ), use the substitution ( u = \frac{1}{x} ). Then ( du = -\frac{1}{x^2}dx ). This transforms the integral into ( -6\int e^u , du ). Integrating ( e^u ) gives ( e^u + C ). Substituting back ( u = \frac{1}{x} ) gives the final result of ( -6e^{1/x} + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.