#intx^2/(x^2+4)dx#
#=int(x^2+4-4)/(x^2+4)dx#
#=int(x^2+4)/(x^2+4)-4/(x^2+4)dx#
#=int1-4/(x^2+4)dx = intdx - int4/(x^2+4)dx#
For the second integral substitute: #2tan(u) = x#
#-> 2sec^2(u)du = dx#
#=intdx-4int(2sec^2(u))/(4tan^2(u)+4)du#
#=intdx-2int(sec^2(u))/(tan^2(u)+1)du#
Now: #sin^2(u)+cos^2(u)=1#
Divide by #cos^2(u)# to get:
#-> sin^2(u)/cos^2(u)+cos^2(u)/cos^2(u)=1/cos^2(u)#
#-> tan^2(u)+1=sec^2(u)#
We can simplify our integral to:
#intdx-2intsec^2(u)/sec^2(u)du#
#=intdx-2intdu=x-2u+C#
Reverse the substitution for #u# to get:
#x -2tan^(-1)(x/2)+C#
Edit: Of course if you knew the integral at the step before I did the substitution, i.e. that:
#int1/(x^2+a^2)dx =1/a tan^(-1)(x/a)#
then you can just integrate directly rather than using substitution.
