To solve this problem we should use the formula
#L=int_a^b sqrt(1+(f^'(x)^2))*dx#
#f(x)=sqrt(x+3)#
#f^'(x)=(1/2)(x+3)^(-1/2)#
#f^'(x)=(1/2)(x+3)^(-1)#
So we have the indefinite integral that solves the problem:
#F(x)=int sqrt (1-1/4(1/(x+3)))*dx=1/2int sqrt(4-(1/(x+3)))*dx#
Lets begin by using a trigonometrical substitution:
#(1/(x+3))=4*tan^2y#
#-dx/(x+3)^2*dx=4*2tanysec^2y*dy#
#-> -16tan^4y*dx=8tanysec^2y*dy# => #dx=-(sec^2y)/(2tan^3y)dy#
Then the indefinite integral becomes
#1/2int cancel(2)secy(-(sec^2y)/(cancel(2)tan^3y))dy=-1/2int sec^3y/(tan^3y)dy#
#=-1/2int 1/cancel(cos^3y)(cancel(cos^3y)/sin^3y)*dy=-1/2 int csc^3y*dy#
We find this last integral in the more complete tables. Anyway, I'll solve it:
#-1/2int csc^3y*dy=-1/2int cscy*dy-1/2int cscy*cot^2y*dy#
I call this last result as expression 1, and I go on to solve its last term:
#int cscy*cot^2y*dy=int (1/(siny))(cos^2y/sin^2y)dy#
Making
#sin y=z#
#cosy*dy=dz#
So
#=int sqrt(1-z^2)/z^3dz#
Using the rule
#int udv=uv-int vdu#
#-> u=(1-z^2)^(1/2)# => #du=-2z(1/2)(1-z^2)^(-1/2)=-z(1-z^2)^(-1/2)#
#-> dv=dz/z^3# => #v=-1/(2z^2)#
we get
#=-(1-z^2)^(1/2)/(2z^2)-1/2int dz/(z*sqrt(1-z^2))dz#
Calling the result above as expression 2 and proceeding to solve its last term
#-1/2int dz/(z*sqrt(1-z^2))dz=#
#z=sin alpha#
#dz=cos alpha*dalpha#
Therefore
#=-1/2int cancel(cos alpha)/(sin alpha*cancel(cos alpha))dalpha=1/2 int csc alpha*dalpha#
#=-1/2*ln|csc alpha-cot alpha|=-1/2*ln|1/z-sqrt(1-z^2)/z|=-1/2*ln|(1-sqrt(1-z^2))/z|#
Returning to expression 2 :
#=-(1-z^2)^(1/2)/(2z^2)-1/2*ln|(1-sqrt(1-z^2))/z|#
But #z=siny#
#-> =-cosy/(2sin^2y)-1/2*ln|(1-cosy)/siny|=-1/2coty*cscy-1/2*ln|cscy-coty|#
Returning to expression 1 ::
#=-1/2ln|cscy-cot y|+1/4*coty*cscy+1/4*ln|cscy-coty|#
#=-1/4ln|cscy-cot y|+1/4*coty*cscy#
Since
#tan^2y=1/(4(x+3))#
we find that
#cscy=sqrt(4x+13); cot y=2sqrt(x+3);siny=1/sqrt(4x+13) and cosy=2sqrt((x+3)/(4x+13))#
So
#F(x)=-1/4*ln|sqrt(4x+13)-2sqrt(x+3)|+(1/4)2sqrt(x+3)*sqrt(4x+13)#
#F(x)= -1/4*ln|sqrt(4x+13)-2sqrt(x+3)|+1/2*sqrt((x+3)(4x+13))#
Finally
#L=F(x=3)-F(x=1)#
#L=-ln(sqrt 25-2sqrt6)/4+sqrt(6*25)/2-((-ln(sqrt(17)-2sqrt4))/4+sqrt(4*17)/2)#
#L=-1/4ln((5-2sqrt16)/(-4+sqrt(17)))+5/2sqrt6-sqrt17~=2.0500#
