As (x^3+1)/((x^2-4)(x^2+1))=(x^3+1)/((x-2)(x+2)(x^2+1))x3+1(x2−4)(x2+1)=x3+1(x−2)(x+2)(x2+1)
Let (x^3+1)/((x^2-4)(x^2+1))=A/(x-2)+B/(x+2)+(Cx+D)/(x^2+1)x3+1(x2−4)(x2+1)=Ax−2+Bx+2+Cx+Dx2+1
or (x^3+1)=A(x+2)(x^2+1)+B(x-2)(x^2+1)+(Cx+D)(x+2)(x-2)(x3+1)=A(x+2)(x2+1)+B(x−2)(x2+1)+(Cx+D)(x+2)(x−2)
when x=2x=2 we have 9=20A9=20A i.e. A=9/20A=920
when x=-2x=−2 we have -7=-20B−7=−20B i.e. B=7/20B=720
Comparing coefficients of x^3x3 and constant term we get
1=A+B+C1=A+B+C i.e. C=1-A-B=1-9/20-7/20=4/20=1/5C=1−A−B=1−920−720=420=15
and 1=2A-2B-4D1=2A−2B−4D i.e. D=-(1-18/20+14/20)/4=1/5D=−1−1820+14204=15
Hence (x^3+1)/((x^2-4)(x^2+1))=9/(20(x-2))+7/(20(x+2))+(x+1)/(5(x^2+1))x3+1(x2−4)(x2+1)=920(x−2)+720(x+2)+x+15(x2+1) and
int(x^3+1)/((x^2-4)(x^2+1))dx∫x3+1(x2−4)(x2+1)dx
= int9/(20(x-2))dx+int7/(20(x+2))dx+int(x+1)/(5(x^2+1))dx∫920(x−2)dx+∫720(x+2)dx+∫x+15(x2+1)dx
= 9/20ln|x-2|+7/20ln|x+2|+1/5(1/2int(2x)/(x^2+1)dx+int1/(x^2+1)dx)920ln|x−2|+720ln|x+2|+15(12∫2xx2+1dx+∫1x2+1dx)
= 9/20ln|x-2|+7/20ln|x+2|+1/10ln|x^2+1|+1/5tan^(-1)x920ln|x−2|+720ln|x+2|+110ln∣∣x2+1∣∣+15tan−1x
