Partial fractions of #(x^2+2x)/(x^2+1)^2#, will be of type
#(x^2+2x)/(x^2+1)^2=(Ax+B)/(x^2+1)+(Cx+D)/(x^2+1)^2#
= #((Ax+B)(x^2+1)+Cx+D)/(x^2+1)^2#
= #(Ax^3+Bx^2+Ax+B+Cx+D)/(x^2+1)^2#
= #(Ax^3+Bx^2+(A+C)x+(B+D))/(x^2+1)^2#
Comparing the terms, we get #A=0#, #B=1#, #A+C=2# and #B+D=0#
i.e. #A=0#, #B=1#, #C=2# and #D=-1#
Hence, #(x^2+2x)/(x^2+1)^2=1/(x^2+1)+(2x-1)/(x^2+1)^2# and
#int(x^2+2x)/(x^2+1)^2dx=int1/(x^2+1)dx+int(2x-1)/(x^2+1)^2dx#
Now #int1/(x^2+1)dx=tan^(-1)x# is a well known integral and for the other one
We can break #int(2x-1)/(x^2+1)^2dx=int(2x)/(x^2+1)^2dx-int1/(x^2+1)^2dx#
For first term, let us assume #u=x^2+1#, then #du=2xdx#= #int(2x)/(x^2+1)^2dx=int(du)/u^2=-1/u=-1/(x^2+1)#
and for second term let us assume #x=tanv# then
#int1/(x^2+1)^2dx=int(sec^2v)/sec^4vdv=intcos^2vdv#
= #1/2int(1+cos2v)dv=v/2+1/4sin2v#
= #tan^(-1)x/2+1/2sinvcosv#
= #tan^(-1)x/2+1/2x/sqrt(x^2+1)xx1/sqrt(x^2+1)#
= #tan^(-1)x/2+1/2x/(x^2+1)#
Hence, #int(2x-1)/(x^2+1)^2dx#
= #-1/(x^2+1)+tan^(-1)x/2+1/2x/(x^2+1)+c#
= #1/2(tan^(-1)x)+(x-2)/(x^2+1)+c#
