Question

$\int\limits { \frac{-8+x}{-16-12x+x^3} } \, dx$

Answer 1

-84x/(x+2)²(x-4)=A/(x+2) +B/(x+2)²+C/(x-4)
A(x+2)(x-4)+B(x-4)+C(x+2)²=-84x
AX²-2AX-8A+Bx-4B+Cx²+4Cx+4C=-84x
x²(A+C)+x(-2A+B+4C)+(-8A-4B+4C)=-84x
A+C=0⇒A=-C
-8A-4B+4C=0⇒8C-4B+4C=0⇒4B=12C⇒B=3C
-2A+B+4C=-84⇒2C+3C=4C=-84⇒9C=-84⇒C=-84/9=-28/3
A=28/3  B=28    C=-28/3
$\int\limits {-84x/(x^3-12x-16)} \, dx =28/3 \int\limits{1/(x+2)} \, dx -28 \int\limits {1/(x+2)^2} \, dx -$$28/3 \int\limits {1/(x-4)} \, dx =28/3*ln(x+2)+28/(x+2)-28/3*ln(x-4)$$28/3*ln[(x+2)/(x-4)]-28/(x+2)+C$