The formula for arc length of a curve defined by #f(x)# is:
#s=int_a^bsqrt(1+f'(x)^2)#
First let's find the derivative #f'(x)#:
#f(x)=x+x(x+3)^(1/2)#
#f'(x)=1+(x+3)^(1/2)+1/2x(x+3)^(-1/2)# (by the product rule)
#f'(x)=1+sqrt(x+3)+x/(2sqrt(x+3))#
Now let's evaluate #f'(x)^2#
#f'(x)^2=[1+sqrt(x+3)+x/(2sqrt(x+3))]^2#
#=1+sqrt(x+3)+x/(2sqrt(x+3))+sqrt(x+3)+(x+3)+(xcancel(sqrt(x+3)))/(2cancel(sqrt(x+3)))+x/(2sqrt(x+3))+(xcancel(sqrt(x+3)))/(2cancel(sqrt(x+3)))+x^2/(4(x+3))#
#=1+2sqrt(x+3)+cancel(2)x/(cancel(2)sqrt(x+3))+(x+3)+cancel(2)(x/cancel(2))+x^2/(4(x+3))#
#=x^2/(4(x+3))+x(2+1/sqrt(x+3))+2sqrt(x+3)+4#
Now you would need to evaluate this (very nasty) integral:
#int_-3^0sqrt(x^2/(4(x+3))+x(2+1/sqrt(x+3))+2sqrt(x+3)+5)#
I punched it into Wolfram|Alpha and got the following result:
So the answer is
#s~~6.54696#
