I've presented two methods, the first is using partial fractions and the second is using substitution of hyperbolic functions.
First, partial fractions.
#intdx/(9x^2-16)#
Split into partial fractions:
#1/(9x^2-16)=1/((3x+4)(3x-4))=A/(3x+4)+B/(3x-4)#
Multiply through by #(3x-4)(3x+4)# to get:
#1=A(3x-4)+B(3x+4)#
To find #A# let #x=-4/3# to cancel the #B# term:
#1=A(3(-4/3)-4)+0=-8A-> A=-1/8#
To find #B# let #x=4/3# to cancel the #A# term:
#1=0+ B(3(4/3)+4)=8B-> B=1/8#
Hence:
#1/((3x+4)(3x-4))=1/(8(3x-4))-1/(8(3x+4))#
So:
#intdx/(9x^2-16)=int1/(8(3x-4))-1/(8(3x+4))dx#
#=1/24ln(3x-4)-1/24ln(3x+4)+C#
An alternative route which involves hyperbolic function substitution:
#intdx/(9x^2-16)#
Let: #3x = 4tanhu->3dx=4sech^2udu#
The substitution now becomes:
#int(4/3sech^2u)/(16tanh^2(u)-16)du=((4/3))/16int(sech^2u)/(tanh^2(u)-1)du#
#-1/12int(sech^2u)/(1-tanh^2(u))du#
Use the identity:
#cosh^2u-sinh^2u=1# (Divide by #cosh^2u#)
#->1-tanh^2u=sech^2u#
So:
#-1/12int(sech^2u)/(1-tanh^2(u))du=-1/12intsech^2u/sech^2udu#
#=-1/12intdu=-1/12u+C#
Reverse the substitution:
#=-1/12tanh^(-1)(3/4x)+C#
By definition we know that: #tanh^-1y = 1/2ln((1+y)/(1-y)) #
Putting this into our answer we get:
#=-1/24ln((1+3/4x)/(1-3/4)x)+C=-1/24ln(((4+3x)/4)/((4-3x)/4))+C#
Cancelling the #4#s.
#-1/24ln(((4+3x))/((4-3x)))+C=1/24ln(((4-3x))/((4+3x)))+C#
In the last line the negative was put into the logarithm to take the reciprocal of the value inside it, now break it up using the laws of logs to get the same answer as the previous method:
#=1/24ln(4-3x)-1/24ln(4+3x)+C#
