The arclength formula is
#L=int_a^b sqrt(1+f'(x)) dx#
First, find the derivative of the #f(x)#
#d/dx(sqrt((x+3)(x/2-1))+5x)#
#d/dx(sqrt((x+3)(x/2-1)))+d/dx(5x)#
#d/dx(sqrt((x+3)(x/2-1)))+5#
For the first term, let #u=(x+3)(x/2-1)=1/2 x^2+1/2 x -3#
#(du)/dx =x+1/2#
Using #u#-substitution, find #d/dx(sqrt((x+3)(x/2-1)))#
#d/dx(sqrt(u))=1/(2sqrt(u)) (du)/dx#
Replacing these values with #x#, gives
#d/dx(sqrt(u))=(x+1/2)/(2sqrt(1/2 x^2+1/2 x -3))#
Plug this result into the arc-length formula above
#L=int_a^b sqrt(1+f'(x)) dx#
#L=int_6^7 sqrt(1+(x+1/2)/(2sqrt(1/2 x^2+1/2 x -3))) dx#
This is a very difficult function to integrate, so using numeric methods is recommended. Using Simpson's Rule (with #n=10# and #Deltax=1//10#), you could calculate
#L~~(1//10)/3[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+cdots#
#cdots+4f(x_10)+f(x_11)]~~1.33#
