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

Mar 15, 2015

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

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

Let's assume:

${e}^{3 x} = 6 \cosh t \Rightarrow 3 x = \ln 6 \cosh t \Rightarrow x = \frac{1}{3} \ln 6 \cosh t \Rightarrow$

$\mathrm{dx} = \frac{1}{3} \cdot \frac{6 \sinh t}{6 \cosh t} \mathrm{dt} = \frac{1}{3} \sinh \frac{t}{\cosh} t \mathrm{dt}$.

Our integral becomes:

$\int \frac{6 \cosh t}{\sqrt{36 {\cosh}^{2} t - 36}} \cdot \frac{1}{3} \sinh \frac{t}{\cosh} t \mathrm{dt} = 2 \int \sinh \frac{t}{6 \sqrt{{\cosh}^{2} t - 1}} \mathrm{dt} =$

$= \frac{1}{3} \int \sinh \frac{t}{\sinh} t \mathrm{dt} = \frac{1}{3} \int \mathrm{dt} = \frac{1}{3} t + c = \left(1\right)$.

Since ${e}^{3 x} = 6 \cosh t \Rightarrow \cosh t = {e}^{3 x} / 6 \Rightarrow t = \arccos h \left({e}^{3 x} / 6\right)$.

So:

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

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

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

So:

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

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

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

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

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