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How do you integrate #int e^sqrtx/sqrtxdx#?

Answer 1

Christopher Agresta

Is an inmediate integral. See below

If we make a change of variable is more evident

Lets do #t=x^2# then #2tdt=dx# and integral becomes

#inte^t/cancelt·2canceltdt=2e^t=2e^sqrtx+C#

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Answer 2

Charles Anctil

To integrate ( \int e^{\sqrt{x}} / \sqrt{x} , dx ), use the substitution ( u = \sqrt{x} ). Then, ( dx = 2u , du ). The integral becomes ( \int 2e^u , du ). Integrating this gives ( 2e^u + C ), where ( C ) is the constant of integration. Substituting back ( u = \sqrt{x} ), the final result is ( 2e^{\sqrt{x}} + C ).

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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.