Solve `int 8/(16-x^2)dx`

Factor the denominator and pull the `8` outside the integral.

`=8int 1/((4-x)(x+4))dx`

Preform partial fraction decomposition on `1/((4-x)(x+4))` .

`1/((4-x)(x+4))=A/(4-x)+B/(x+4)`

`1=A(x+4)+B(x-4)`

`1=Ax+4A+Bx-4B`

`1=(A+B)x+4(A-B)`

Sine the left hand coefficients must be equal to the right hand side coefficients, `(A+B)` must be equal to zero to make `x` vanish and `4(A-B)` must equal `1` .

`A+B=0`

`A=-B`

`1=4(A-B)`

`1/4=A-B`

`1/4=2A`

`1/8=A, -1/8=B`

Then the integral becomes:

`=8int (1/8)[1/(4-x) -1/(x+4)]dx`

`=int 1/(4-x)dx-int 1/(x+4)dx`

Use u-substitution on the first integral.

`4-x=u` , and `du=-dx`

on the 2nd integral.

`x+4=v` , `dv=dx`

`=-int 1/(u)du-int 1/(v)dv`

`=-ln|u|-ln|v|+C`

`=-ln|4-x|-ln|x+4|+C`

`=-ln((4-x)/(x+4))+C`

`=ln(((4-x)/(x+4))^-1)+C`

Then finally,

`int 8/(16-x^2)dx=ln((x+4)/(4-x))+C`