The solution is 1/72(ln|(e^(3x)-6)/(e^(3x)+6)|)+c172(ln∣∣∣e3x−6e3x+6∣∣∣)+c.
First of all the 1/212 can go out of the sign of integral and it is possible to notice that e^(6x)=(e^(3x))^2e6x=(e3x)2.
Then the substitution method will be used:
e^(3x)=trArr3x=ln(t)rArrx=1/3ln(t)rArrdx=(1/3)(1/t)dte3x=t⇒3x=ln(t)⇒x=13ln(t)⇒dx=(13)(1t)dt
The integral now is:
1/2intt/(t^2-36)(1/3)(1/t)dt=1/6int1/((t-6)(t+6))dt12∫tt2−36(13)(1t)dt=16∫1(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))1(t−6)(t+6)=At−6+Bt+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 AA and BB have to be find.
The numerator 11 has to be identical to A(t+6)+B(t-6)A(t+6)+B(t−6).
Two polynomials are identical if they assume the same values for the same value of xx.
So:
If t=6t=6 then 1=A(6+6)+B(6-6)rArr1=12ArArrA=1/121=A(6+6)+B(6−6)⇒1=12A⇒A=112
If t=-6t=−6 then 1=A(-6+6)+B(-6-6)rArr1=-12BrArrB=-1/121=A(−6+6)+B(−6−6)⇒1=−12B⇒B=−112
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|)+c16∫(112t−6−112t+6)dt=(16)(112)∫(1t−6−1t+6)dt=172(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.
