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

Question

Ezekiel Alexander · Accepted Answer

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#.