How do you integrate #e^(3x)/(e^6x-36)^(1/2)dx#?

Answer 1
The answer is: #1/3ln(e^(3x)+sqrt(e^(6x)-36))+c#.
First of all I assume that there is an error in your writing: I think that #e^6x# would be #e^(6x)#.

Let's assume:

#e^(3x)=6coshtrArr3x=ln6coshtrArrx=1/3ln6coshtrArr#
#dx=1/3*(6sinht)/(6cosht)dt=1/3sinht/coshtdt#.

Our integral becomes:

#int(6cosht)/sqrt(36cosh^2t-36)*1/3sinht/coshtdt=2intsinht/(6sqrt(cosh^2t-1))dt=#
#=1/3intsinht/sinhtdt=1/3intdt=1/3t+c=(1)#.
Since #e^(3x)=6coshtrArrcosht=e^(3x)/6rArrt=arccosh(e^(3x)/6)#.

So:

#(1)=1/3arccosh(e^(3x)/6)+c#.
There is another way to write the solution, remembering the logarithmic expression of the function #y=arccoshx#, that is:
#y=ln(x+sqrt(x^2-1))#.

So:

#(1)=1/3ln(e^(3x)/6+sqrt(e^(6x)/36-1))+c=#
#=1/3ln((e^(3x)+sqrt(e^(6x)-36))/6)+c=#
#=1/3ln(e^(3x)+sqrt(e^(6x)-36))-1/3ln6+c=#
#=1/3ln(e^(3x)+sqrt(e^(6x)-36))+c#
because #-1/3ln6# is a number.
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Answer 2

To integrate the given function (\int \frac{e^{3x}}{\sqrt{e^{6x}-36}} dx), use a substitution method:

  1. Substitution: Let (u = e^{6x} - 36). Then, differentiate both sides with respect to (x) to find (du).

    • (du = 6e^{6x}dx)

    This implies we need a factor of (e^{6x}) in the numerator for the substitution to work directly. Notice that we can rewrite the original integral in a form that makes use of this:

    • Given: (\frac{e^{3x}}{\sqrt{e^{6x}-36}} dx)

      We rewrite (e^{3x}) as (\frac{e^{6x}}{e^{3x}}) to get: (\frac{\frac{e^{6x}}{e^{3x}}}{\sqrt{e^{6x}-36}} dx)

  2. Adjust for (du): We know that (du = 6e^{6x}dx), but we have (\frac{e^{6x}}{e^{3x}}) in our integral, not (6e^{6x}). To compensate, we multiply and divide by 6:

    • (\int \frac{1}{6} \cdot \frac{6e^{6x}}{e^{3x} \sqrt{e^{6x}-36}} dx)

      So, we adjust the integral to fit our (du), giving us: (\frac{1}{6} \int \frac{6e^{6x}}{e^{3x} \sqrt{e^{6x}-36}} dx)

  3. Substitute and Integrate: Now, substitute (u = e^{6x} - 36) and (du = 6e^{6x}dx):

    • (\frac{1}{6} \int \frac{du}{\sqrt{u}})
  4. Integrate with Respect to (u):

    • The integral of (\frac{1}{\sqrt{u}}) is (2\sqrt{u}), so we get (\frac{1}{6} \cdot 2\sqrt{u} + C) or (\frac{1}{3}\sqrt{u} + C).
  5. Substitute Back: Replace (u) with (e^{6x} - 36):

    • (\frac{1}{3}\sqrt{e^{6x} - 36} + C)

Thus, the integral (\int \frac{e^{3x}}{\sqrt{e^{6x}-36}} dx) is (\frac{1}{3}\sqrt{e^{6x} - 36} + 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.

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.

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.

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.

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