Substitute #x=u^2#. Then #dx = 2udu# and the integral becomes
#int_0^3dx/((x+2)(sqrtx+1) ) = int_0^{sqrt3} (2udu)/((u^2+2)(u+1)) #
We will now evaluate this integral using partial fractions. to thsi end, let us start with
# (2u)/((u^2+2)(u+1)) = A/(u+1)+(Bu+C)/(u^2+2)qquad implies#
#2u = A(u^2+2)+(Bu+C)(u+1) = (A+B)u^2+(B+C)u+2A+C#
Comparing coefficients on both sides we get
#A+B = 0,qquad B+C = 2, qquad 2A+C = 0#
This can be easily solved to get
#A = -2/3,quad B = 2/3, quad"and " C=4/3#
Thus
# (2u)/((u^2+2)(u+1)) = -2/3 1/(u+1)+1/3 (2u+4)/(u^2+2)#
and so
#int (2udu)/((u^2+2)(u+1)) = -2/3 int (du)/(u+1)+1/3 int ((2u+4)du)/(u^2+2)#
#qquad = -2/3 ln(u+1)+1/3 ln(u^2+2)+4/(3sqrt2) tan^-1(u/sqrt2)+C#
Thus, the definite integral is
#( -2/3 ln(u+1)+1/3 ln(u^2+2)+4/(3sqrt2) tan^-1(u/sqrt2))_0^{sqrt3} #
#= -2/3ln(1+sqrt3)+1/3ln(5/2) +2sqrt2/3 tan^-1(sqrt(3/2))#
