Question

Question #639a7

Answer 1

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

Explanation:

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`

Answer 2

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.