Question #639a7

Factorize the denominator:

#1/(9x^2-16) = 1/((3x+4)(3x-4))#

Then use partial fractions:

#1/(9x^2-16) = A/(3x+4)+B/(3x-4)#

#1/(9x^2-16) = (A(3x-4)+B(3x+4))/((3x+4)(3x-4))#

#1/(9x^2-16) = (3Ax+3Bx-4A+4B)/((3x+4)(3x-4))#

#{(3A+3B=0),(-4A+4B=1):}#

#{(A=-B),(A-B=-1/4):}#

#{(A=-1/8),(B=1/8):}#

So:

#int dx/(9x^2-16) = 1/8 int dx/(3x-4) - 1/8 int dx/(3x+4)#

#int dx/(9x^2-16) = 1/24 int (d(3x-4))/(3x-4) - 1/24 int (d(3x+4))/(3x+4)#

#int dx/(9x^2-16) = 1/24 (ln abs (3x-4) - ln abs (3x+4)) +C#

and using the properties of logarithms:

#int dx/(9x^2-16) = 1/24 ln (abs (3x-4)/abs (3x+4)) +C#

Your method can work.

#intdx/(9x^2-16)#

#3x=4sectheta# so #3dx=4secthetatanthetad theta#:

#=int(4/3secthetatanthetad theta)/(16sec^2theta-16)d theta=1/12int(secthetatantheta)/tan^2thetad theta=1/12intcscthetad theta#

#=1/12lnabs(csctheta-cottheta)#

Which can be rewritten as:

#=1/12lnabs(sectheta/tantheta-1/tantheta)=1/12lnabs((sectheta-1)/tantheta)#

Then using #sectheta=3/4x# and #tantheta=sqrt(sec^2theta-1)=sqrt(9/16x^2-1)=1/4sqrt(9x^2-16)#:

#=1/12lnabs((3/4x-1)/(1/4sqrt(9x^2-16)))#

#=1/12lnabs((3x-4)/sqrt(9x^2-16))#

This is where the "simplification" gets tricky:

#=1/12lnabs(sqrt((3x-4)^2/(9x^2-16)))#

Bringing the square root out of the logarithm as a #1//2# power using #log(a^b)=blog(a)#:

#=1/24lnabs(((3x-4)^2)/(9x^2-16))#

#=1/24lnabs((3x-4)^2/((3x+4)(3x-4)))#

#=1/24lnabs((3x-4)/(3x+4))+C#

Which is the answer found through using partial fractions. As much as I love trig substitutions, there are definitely times when using partial fractions will get you to the simplest answer the fastest.