# How do I evaluate int(e^(3x))/(e^(6x)-36)1/2dx?

Jan 28, 2015

The solution is $\frac{1}{72} \left(\ln | \frac{{e}^{3 x} - 6}{{e}^{3 x} + 6} |\right) + c$.

First of all the $\frac{1}{2}$ can go out of the sign of integral and it is possible to notice that ${e}^{6 x} = {\left({e}^{3 x}\right)}^{2}$.

Then the substitution method will be used:

${e}^{3 x} = t \Rightarrow 3 x = \ln \left(t\right) \Rightarrow x = \frac{1}{3} \ln \left(t\right) \Rightarrow \mathrm{dx} = \left(\frac{1}{3}\right) \left(\frac{1}{t}\right) \mathrm{dt}$

The integral now is:

$\frac{1}{2} \int \frac{t}{{t}^{2} - 36} \left(\frac{1}{3}\right) \left(\frac{1}{t}\right) \mathrm{dt} = \frac{1}{6} \int \frac{1}{\left(t - 6\right) \left(t + 6\right)} \mathrm{dt}$

In the last passage the denominator was factored.

Now it is necessary to "disjoint" the fraction in two fractions using this method:

$\frac{1}{\left(t - 6\right) \left(t + 6\right)} = \frac{A}{t - 6} + \frac{B}{t + 6} = \frac{A \left(t + 6\right) + B \left(t - 6\right)}{\left(t - 6\right) \left(t + 6\right)}$

The fraction on the left has to be identical to the one on the right, the two denominators are identical, so $A$ and $B$ have to be find.

The numerator $1$ has to be identical to $A \left(t + 6\right) + B \left(t - 6\right)$.
Two polynomials are identical if they assume the same values for the same value of $x$.

So:

If $t = 6$ then $1 = A \left(6 + 6\right) + B \left(6 - 6\right) \Rightarrow 1 = 12 A \Rightarrow A = \frac{1}{12}$

If $t = - 6$ then $1 = A \left(- 6 + 6\right) + B \left(- 6 - 6\right) \Rightarrow 1 = - 12 B \Rightarrow B = - \frac{1}{12}$

The integral bacames:

$\frac{1}{6} \int \left(\frac{\frac{1}{12}}{t - 6} - \frac{\frac{1}{12}}{t + 6}\right) \mathrm{dt} = \left(\frac{1}{6}\right) \left(\frac{1}{12}\right) \int \left(\frac{1}{t - 6} - \frac{1}{t + 6}\right) \mathrm{dt} = \frac{1}{72} \left(\ln | t - 6 | - \ln | t + 6 |\right) + c$

In the last passage it was used the immediate integral:

$\int \frac{f ' \left(x\right)}{f} \left(x\right) \mathrm{dx} = \ln | f \left(x\right) | + c$.

It not necessary, but kind, to use the logarithmic property: $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$.

So the solution is: $\frac{1}{72} \ln | \frac{t - 6}{t + 6} | + c$.

Now it is necessary to "return" to the old variable, $x$.

($t = {e}^{3 x}$)

$I = \frac{1}{72} \left(\ln | \frac{{e}^{3 x} - 6}{{e}^{3 x} + 6} |\right) + c$.