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Mathematics Question on integral

Integrate the rational function: (x2+1)(x2+2)(x2+3)(x2+4)\frac {(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}

Answer

(x2+1)(x2+2)(x2+3)(x2+4)\frac {(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)} = 1−(4x2+10)(x2+3)(x2+4)1- \frac {(4x^2+10)}{(x^2+3)(x^2+4)}

Let (4x2+10)(x2+3)(x2+4)\frac {(4x^2+10)}{(x^2+3)(x^2+4)} = Ax+B(x2+3)+Cx+D(x2+4)\frac {Ax+B}{(x^2+3)}+\frac {Cx+D}{(x^2+4)}

4x2+10=(Ax+B)(x2+4)+(Cx+D)(x2+3)4x^2+10 = (Ax+B)(x^2+4)+(Cx+D)(x^2+3)

4x2+10=Ax3+4Ax+Bx2+4B+Cx3+3Cx+Dx2+3D4x^2+10 = Ax^3+4Ax+Bx^2+4B+Cx^3+3Cx+Dx^2+3D

4x2+10=(A+C)x3+(B+D)x2+(4A+3C)x+(4B+3D)4x^2+10 = (A+C)x^3+(B+D)x^2+(4A+3C)x+(4B+3D)

Equating the coefficients of x3,x2,x, and constant term, we obtainEquating\ the\ coefficients\ of\ x^3, x^2, x,\ and\ constant \ term,\ we\ obtain
A+C=0A + C = 0
B+D=4B + D = 4
4A+3C=04A + 3C = 0
4B+3D=104B + 3D = 10
On solving these equations, we obtainOn\ solving\ these \ equations,\ we \ obtain
A=0,B=−2,C=0, and D=6A = 0, B = −2, C = 0, \ and\ D = 6

∴ (4x2+10)(x2+3)(x2+4)\frac {(4x^2+10)}{(x^2+3)(x^2+4)} =−2(x2+3)+6(x2+4)-\frac {2}{(x^2+3)}+\frac {6}{(x^2+4)}

(x2+1)(x2+2)(x2+3)(x2+4)\frac {(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)} = 1−2(x2+3)+6(x2+4)-\frac {2}{(x^2+3)}+\frac {6}{(x^2+4)}

⇒ ∫$$\frac {(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\ dx = ∫[1+2(x2+3)−6(x2+4)]dx∫[{1+\frac {2}{(x^2+3)}-\frac {6}{(x^2+4)}}]dx

                                  = $∫{1+\frac {2}{x^2+(\sqrt 3)^2}-\frac {6}{x^2+2^2}}dx$

                                  = $x+2(\frac {1}{\sqrt 3}tan^{-1}\frac {x}{\sqrt 3})-6(\frac 12 tan^{-1}\frac x2)+C$

                                  =  $x+\frac {2}{\sqrt 3}tan^{-1}\frac { x}{\sqrt 3}-3tan^{-1}\frac x2+C$