We want to solve
#I=int1/(x^2+1)^2dx#
We would use Partial Fractions, when the factors in the denominator are different, which they isn't here
If we apply the normal method of partial fraction,
we would end up with our originally problem
#1/(x^2+1)^2=(Ax+B)/(x^2+1)+(Cx+D)/(x^2+1)^2#
#1=(Ax+B)(x^2+1)+Cx+D#
#1=Ax^3+Ax+Bx^2+B+Cx+D#
#color(green)0x^3+color(blue)0x^2+color(red)0x+color(orange)1=color(green)Ax^3+color(blue)Bx^2+color(red)((A+C))x+color(orange)((B+D))#
#A=0#, #B=0#, #C=0#, #D=1#
Therefore we end with the original problem
#(0x+0)/(x^2+1)+(0x+1)/(x^2+1)^2=1/(x^2+1)^2#
To solve the integral use substitution
Let #x=tan(u)=>dx/(du)=sec^2(u)#
#I=int1/(tan^2(u)+1)^2sec^2(u)du#
#=int1/(sec^2(u))^2sec^2(u)du#
#=int1/sec^2(u)du#
#=intcos^2(u)du#
Use the identity #cos^2(u)=1/2+1/2cos(2u)#
#I=1/2intdu+1/2intcos(2u)du#
#=1/2u+1/4sin(2u)+C#
#=1/2(u+1/2sin(2u))+C#
Substitute #u=arctan(x)#
#I=1/2(arctan(x)+1/2sin(2arctan(x)))#
#=1/2(arctan(x)+x/(x^2+1))+C#
