Solve the integral int 8/(16-x^) dx

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
https://www.purplemath.com/modules/partfrac.htm