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

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Answer 1 · 2015-03-15T20:04:52

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.

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