#int ln(x^2+4)dx#
Use Integration by Parts
#int u ##dv#/#dx# = uv - #int v ##du#/#dx#
#int ln(x^2+4)dx# = #int ln(x^2+4).1dx#
#u=ln(x^2+4)#
# du=2x/(x^2+4)dx# using chain rule
#dv=1 #
#v=int(1)dx=x+constant#
using the integration by parts formula
#int ln(x^2+4)dx=xln(x^2+4) - intx(2x/(x^2+4))dx#
#=xln(x^2+4)- int(2x^2/(x^2+4))dx#
#=xln(x^2+4)- 2int((x^2+4 -4)/(x^2+4))dx#
#=xln(x^2+4)- 2int((x^2+4 )/(x^2+4)-4/(x^2+4))dx#
#=xln(x^2+4)- 2int(1-4/(x^2+4))dx#
Now the standard integral #int dx/(a^2+x^2) = 1/a tan^-1(x/a)#
Therefore
#=xln(x^2+4)- 2x+2(4int1/(x^2+2^2))dx# ----equation1
#=xln(x^2+4)- 2x+2(4/2tan^-1(x/2))#
#=xln(x^2+4)- 2x+4tan^-1(x/2)#
If you can't remember the standard integral above then we can integrate this using a substitution
#int4/(x^2+4)#
#x=2tanA#
#dx=2sec^2A#dA##
#=int(2sec^2A)/(tan^2A+1)##dA#
#sec^2A-1 = tan^2A#
#=int(2sec^2A)/(sec^2A)##dA#
#=2A+constant#
#= 2tan^-1(x/2)#
This can be substituted into equation 1
