Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 16
int(8x)/(x^3+x^2-x-1)dx (8x)/(x^3+x^2-x-1)=(8x)/((x^3+x^2)-1(x+1)) =(8x)/((x^2(x+1)-1(x+1))) =(8x)/((x+1)(x^2-1)) =(8x)/((x+1)(x+1)(x-1)) =(8x)/((x-1)(x+1)^2) Now let's form the partial fraction template, (8x)/((x-1)(x+1)^2)=A/(x-1)+B/(x+1)+C/(x+1)^2 Multiply the equation by the denominator, 8x=A(x+1)^2+B(x-1)(x+1)+C(x-1) 8x=A(x^2+2x+1)+B(x^2-1)+C(x-1) 8x=Ax^2+2Ax+A+Bx^2-B+Cx-C 8x=(A+B)x^2+(2A+C)x+A-B-C Comparing the coefficients of the like terms, A+B=0 -----------------(1) 2A+C=8 -----------------(2) A-B-C=0 ---------------(3) From equation 1, B=-A Substitute B in equation 3, A-(-A)-C=0 2A-C=0 ---------------(4) Now add equations 2 and 4, 4A=8 A=8/4 A=2 B=-A=-2 Plug in the value of A in equation 4, 2(2)-C=0 C=4 Plug in the values of A, B and C in the partial fraction template, (8x)/((x-1)(x+1)^2)=2/(x-1)+(-2)/(x+1)+4/(x+1)^2 int(8x)/(x^3+x^2-x-1)dx=int(2/(x-1)-2/(x+1)+4/(x+1)^2)dx Apply the sum rule, =int2/(x-1)dx-int2/(x+1)dx+int4/(x+1)^2dx Take the constant out, =...