Determine the Integral : ∫ [1+(e^-x)^1/2]/(e^x)^1/2 dx

Note that, by the laws of exponents, you can write:
[e^(-x)]^(1/2) = e^(-x/2) and (e^x)^(1/2) = e^(x/2).

So, we have:
∫ {1 + [e^(-x)]^(1/2)}/(e^x)^(1/2) dx
= ∫ [1 + e^(-x/2)]/e^(x/2) dx
= ∫ [e^(-x/2) + e^(-x)] dx, by the laws of exponents
= -2e^(-x/2) - e^(-x) + C, by integrating.

I hope this helps!

... ∫ { [ 1 + √(ℯֿˣ) ] / √(ℯˣ) } dx ....................... Now Split

= ∫ { [ 1 / √(ℯˣ) ] + [ √(ℯֿˣ) / √(ℯˣ) ] } dx

= ∫ √(ℯֿˣ) dx + ∫ √(ℯֿ²ˣ) dx

= ∫ ℯֿˣʹ² dx + ∫ ℯֿˣ dx ................ Now Use : ∫ ℯªˣ dx = ( ℯªˣ ) / (a)

= [ ℯֿˣʹ² / (-1/2) ] + [ - ℯֿˣ ] + C

= -2 ℯֿˣʹ² - ℯֿˣ + C .................................. Ans.
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