We first decompose the Integrand into Partial Fracions using
Heavyside's Method.
Let, for, A,B,C,D,E in RR,
(1):(x^3+3x+2)/{(x^2+1)^2(x+1)}=A/(x+1)+(Bx+C)/(x^2+1)+(Dx+E)/(x^2+1)^2.
Then, A=[(x^3+3x+2)/(x^2+1)^2)]_(x=-1) =(-1-3+2)/((1+1)^2)=-1/2.
:., by (1), (x^3+3x+2)/{(x^2+1)^2(x+1)}+(1/2)/(x+1)=(Bx+C)/(x^2+1)+(Dx+E)/(x^2+1)^2.
Upon simplification, this becomes,
{x^3+3x+2+1/2(x^2+1)^2}/{(x^2+1)^2(x+1)}=(1/2x^4+x^3+x^2+3x+5/2)/{(x^2+1)^2(x+1)}={(x+1)(1/2x^3+1/2x^2+1/2x+5/2)}/{(x^2+1)^2(x+1)}
=(1/2x^3+1/2x^2+1/2x+5/2)/(x^2+1)^2
={(Bx+C)(x^2+1)+Dx+E}/(x^2+1)^2
:.1/2x^3+1/2x^2+1/2x+5/2=Bx^3+Cx^2+(B+D)x+(C+E).
rArr B=1/2, C=1/2, B+D=1/2, C+E=5/2, or,
A=-1/2, B=1/2, C=1/2, D=0, E=2.
Therefore, int(x^3+3x+2)/{(x^2+1)^2(x+1)}dx=I, say
=-1/2int1/(x+1)dx+1/2int(x+1)/(x^2+1)dx+2int1/(x^2+1)^2dx
=-1/2ln|x+1|+1/2intx/(x^2+1)dx+1/2int1/(x^2+1)dx+2I_1,
=-1/2ln|x+1|+1/4int{d/dx(x^2+1)}/(x^2+1)dx+1/2arc tanx+2I_1
=-1/2ln|x+1|+1/4ln(x^2+1)+1/2arc tanx+2I_1, where,
I_1=int1/(x^2+1)^2dx
=intsec^2y/sec^4ydy...[x=tany rArr dx=sec^2ydy]
=intcos^2ydy=int{1/2(1+cos2y)}dy=1/2{y+1/2(sin2y)}
Here, x=tanyrArr y=arc tanx; sin2y=(2tany)/(1+tan^2y)=(2x)/(x^2+1)
:. I_1=1/2arc tanx+x/(2(x^2+1))
Altogether, we have,
I=-1/2ln|x+1|+1/4ln(x^2+1)+1/2arc tanx+2{1/2arc tanx+x/(2(x^2+1))}
=-1/2ln|x+1|+1/4ln(x^2+1)+3/2arc tanx+x/(x^2+1)+C, or,
I=ln[root4{(x^2+1)/(x+1)^2}]+3/2arc tanx+x/(x^2+1)+C.
Enjoy Maths.!
