Calculate the indefinite integral first
Let #u=sqrtx#, #=>#, #du=1/(2sqrtx)dx#
#I=int((x+1)dx)/(x+2sqrtx-3)=int(2u(u^2+1)du)/(u^2+2u-3)#
#(2u(u^2+1))/(u^2+2u-3)=2(u-2)+(2(8u-6))/(u^2+2u-3)#
#=2(u-2)+(4(4u-3))/(u^2+2u-3)#
Perform the decomposition into partial fractions
#((4u-3))/(u^2+2u-3)=(4u-3)/((u-1)(u+3))#
#=A/(u-1)+B/(u+3)#
#=(A(u+3)+B(u-1))/((u-1)(u+3))#
The denominators are the same, compare the numerators
#4u-3=A(u+3)+B(u-1)#
Let #u=1#, #=>#, #1=4A#
let #u=-3#, #=>#, #-15=-4B#
Therefore,
#2(u-2)+(4(4u-3))/(u^2+2u-3)=2(u-2)+1/(u-1)+15/(u+3)#
And finally
#I=2int(u-2)du+int(du)/(u-1)+int(15du)/(u+3)#
#=u^2-4u+ln(u-1)+15ln(u+3)#
#=x-4sqrtx+ln(sqrtx-1)+15ln(sqrtx+3)+C#
And the definite integral is
#int_4^9((x+1)dx)/(x+2sqrtx-3)=#
#=[x-4sqrtx+ln(sqrtx-1)+15ln(sqrtx+3)]_4^9#
#=(9-12+ln2+15ln6)-(4-8+ln1+15ln5)#
#=15ln6-15ln5+ln2+1#
#=4.43#