Integrate e^2x / SQRT (e^2x + 3)

  • Thread starter donleo
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In summary, to integrate e^2x / sqrt(e^2x + 3), you can use the substitution t = e^2x + 3 and dt = 2e^2x dx. This leads to the integral becoming 1/2 (dt/sqrt(t)) which then simplifies to 1/2 (sqrt(t)) or sqrt(e^2x + 3) / 2. However, it is important to note that the power of t should be -1/2 instead of 1/2, leading to the final solution of sqrt(e^2x + 3) / 3.
  • #1
donleo
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Homework Statement



Integrate e^2x / SQRT [(e^2x) + 3)]

Homework Equations





The Attempt at a Solution



i know the solution, is: SQRT [(e^2x) + 3]

but i have no idea why. Please I need help

thank you
 
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  • #2
Have you tried substitution?
 
  • #3
eok20 said:
Have you tried substitution?

yes, t=e^2x + 3 but still nothing
 
  • #4
Ok, if t=e^2x + 3 then dt = 2e^2x, and what does the integral become in terms of t and dt?
 
  • #5
eok20 said:
Ok, if t=e^2x + 3 then dt = 2e^2x, and what does the integral become in terms of t and dt?

i have 1/2 (dt/ sqrt t)

so 1/2 t^(3/2) / (3/2)

then sqrt t / 3

then sqrt (e^2x+3)^3 / 3

its not the solution, it may be sqrt (e^2x+3)
 
  • #6
Close, you have 1/2(dt/sqrt(t)) = 1/2 t^(-1/2) dt. You integrated 1/2 t^(1/2) dt instead
 
  • #7
sqrt t is in the Denominator! So the power of t isn't 1/2, but actually... ?
 
  • #8
eok20 said:
Close, you have 1/2(dt/sqrt(t)) = 1/2 t^(-1/2) dt. You integrated 1/2 t^(1/2) dt instead

its true! lol, it was my mistake, i forgot the change t^1/2 -> t^-1/2

thank you!
 

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The function "Integrate e^2x / SQRT (e^2x + 3)" is an integral function that calculates the area under the curve of the expression e^2x / SQRT (e^2x + 3).

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The integral of the function "Integrate e^2x / SQRT (e^2x + 3)" is (1/2)(e^2x + 3)^(1/2) + C, where C is the constant of integration.

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