To solve this problem we should use the formula
L=int_a^b sqrt(1+(f^'(x)^2))*dx
f(x)=sqrt(x-1)
f^'(x)=(1/2)(x-1)^(-1/2)
f^'(x)=(1/2)(x-1)^(-1)
So we have the indefinite integral that solves the problem:
F(x)=int sqrt (1-1/4(1/(x-1)))*dx=1/2int sqrt(4-(1/(x-1)))*dx
Lets begin by using a trigonometrical substitution:
(1/(x-1))=4*tan^2y
-dx/(x-1)^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-1))
we find that
cscy=sqrt(4x-3); cot y=2sqrt(x-1);siny=1/sqrt(4x-3) and cosy=2sqrt((x-1)/(4x-3))
So
F(x)=-1/4*ln|sqrt(4x-3)-2sqrt(x-1)|+(1/4)2sqrt(x-1)*sqrt(4x-3)
F(x)= -1/4*ln|sqrt(4x-3)-2sqrt(x-1)|+1/2*sqrt((x-1)(4x-3))
Finally
L=F(x=3)-F(x=1)
L=-ln(sqrt 5-2)/4+sqrt5/2~=1.4789