How do you integrate $\frac{e^{3 x}}{{(e^{6} x - 36)}^{\frac{1}{2}}} d x$ $e^(3x)/(e^6x-36)^(1/2)dx$ ?

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How do you integrate $\frac{e^{3 x}}{{(e^{6} x - 36)}^{\frac{1}{2}}} d x$ $e^(3x)/(e^6x-36)^(1/2)dx$ ?

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Answer 1 · 2015-03-15T20:04:52

The answer is: $\frac{1}{3} ln (e^{3 x} + \sqrt{e^{6 x} - 36}) + c$ $1/3ln(e^(3x)+sqrt(e^(6x)-36))+c$ .

First of all I assume that there is an error in your writing: I think that $e^{6} x$ $e^6x$ would be $e^{6 x}$ $e^(6x)$ .

Let's assume:

$e^{3 x} = 6 cosh t \Rightarrow 3 x = ln 6 cosh t \Rightarrow x = \frac{1}{3} ln 6 cosh t \Rightarrow$ $e^(3x)=6coshtrArr3x=ln6coshtrArrx=1/3ln6coshtrArr$

$d x = \frac{1}{3} \cdot \frac{6 sinh t}{6 cosh t} d t = \frac{1}{3} \frac{sinh t}{cosh t} d t$ $dx=1/3*(6sinht)/(6cosht)dt=1/3sinht/coshtdt$ .

Our integral becomes:

$\int \frac{6 cosh t}{\sqrt{36 {cosh}^{2} t - 36}} \cdot \frac{1}{3} \frac{sinh t}{cosh t} d t = 2 \int \frac{sinh t}{6 \sqrt{{cosh}^{2} t - 1}} d t =$ $int(6cosht)/sqrt(36cosh^2t-36)*1/3sinht/coshtdt=2intsinht/(6sqrt(cosh^2t-1))dt=$

$= \frac{1}{3} \int \frac{sinh t}{sinh t} d t = \frac{1}{3} \int d t = \frac{1}{3} t + c = (1)$ $=1/3intsinht/sinhtdt=1/3intdt=1/3t+c=(1)$ .

Since $e^{3 x} = 6 cosh t \Rightarrow cosh t = \frac{e^{3 x}}{6} \Rightarrow t = arccos h (\frac{e^{3 x}}{6})$ $e^(3x)=6coshtrArrcosht=e^(3x)/6rArrt=arccosh(e^(3x)/6)$ .

So:

$(1) = \frac{1}{3} arccos h (\frac{e^{3 x}}{6}) + c$ $(1)=1/3arccosh(e^(3x)/6)+c$ .

There is another way to write the solution, remembering the logarithmic expression of the function $y = arccos h x$ $y=arccoshx$ , that is:

$y = ln (x + \sqrt{x^{2} - 1})$ $y=ln(x+sqrt(x^2-1))$ .

So:

$(1) = \frac{1}{3} ln (\frac{e^{3 x}}{6} + \sqrt{\frac{e^{6 x}}{36} - 1}) + c =$ $(1)=1/3ln(e^(3x)/6+sqrt(e^(6x)/36-1))+c=$

$= \frac{1}{3} ln (\frac{e^{3 x} + \sqrt{e^{6 x} - 36}}{6}) + c =$ $=1/3ln((e^(3x)+sqrt(e^(6x)-36))/6)+c=$

$= \frac{1}{3} ln (e^{3 x} + \sqrt{e^{6 x} - 36}) - \frac{1}{3} ln 6 + c =$ $=1/3ln(e^(3x)+sqrt(e^(6x)-36))-1/3ln6+c=$

$= \frac{1}{3} ln (e^{3 x} + \sqrt{e^{6 x} - 36}) + c$ $=1/3ln(e^(3x)+sqrt(e^(6x)-36))+c$

because $- \frac{1}{3} ln 6$ $-1/3ln6$ is a number.

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