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

The solution is `1/72(ln|(e^(3x)-6)/(e^(3x)+6)|)+c`.

First of all the `1/2` can go out of the sign of integral and it is possible to notice that `e^(6x)=(e^(3x))^2`.

Then the substitution method will be used:

`e^(3x)=trArr3x=ln(t)rArrx=1/3ln(t)rArrdx=(1/3)(1/t)dt`

The integral now is:

`1/2intt/(t^2-36)(1/3)(1/t)dt=1/6int1/((t-6)(t+6))dt`

In the last passage the denominator was factored.

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

`1/((t-6)(t+6))=A/(t-6)+B/(t+6)=(A(t+6)+B(t-6))/((t-6)(t+6))`

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(t+6)+B(t-6)`.
Two polynomials are identical if they assume the same values for the same value of `x`.

So:

If `t=6` then `1=A(6+6)+B(6-6)rArr1=12ArArrA=1/12`

If `t=-6` then `1=A(-6+6)+B(-6-6)rArr1=-12BrArrB=-1/12`

The integral bacames:

`1/6int((1/12)/(t-6)-(1/12)/(t+6))dt=(1/6)(1/12)int(1/(t-6)-1/(t+6))dt=1/72(ln|t-6|-ln|t+6|)+c`

In the last passage it was used the immediate integral:

`int(f'(x))/f(x)dx=ln|f(x)|+c`.

It not necessary, but kind, to use the logarithmic property: `lna-lnb=ln(a/b)`.

So the solution is: `1/72ln|(t-6)/(t+6)|+c`.

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

(`t=e^(3x)`)

`I=1/72(ln|(e^(3x)-6)/(e^(3x)+6)|)+c`.