Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 24

int1/sqrt(x^2-4)dx
Let's apply integral substitution:x=2sec(u)
=>dx=2sec(u)tan(u)du
=int1/sqrt((2sec(u))^2-4)2sec(u)tan(u)du
=int(2sec(u)tan(u))/sqrt(4sec^2(u)-4)du
=int(2secutan(u))/(sqrt(4)sqrt(sec^2(u)-1))du
Now use the trigonometric identity: tan^2(x)=sec^2(x)-1
=int(2sec(u)tan(u))/(2sqrt(tan^2(u)))du
=intsec(u)du
Now use the standard integral:intsec(x)dx=ln|sec(x)+tan(x)|
=ln|sec(u)+tan(u)| ----------(1)
Now from the substitution:
sec(u)=x/2 and,
tan^2(u)=sec^2(u)-1
tan^2(u)=(x/2)^2-1
tan^2(u)=(x^2-4)/4
tan(u)=sqrt(x^2-4)/2
Substitute back the above in the result (1)
=ln|x/2+sqrt(x^2-4)/2|
=ln|(x+sqrt(x^2-4))/2|
=ln|x+sqrt(x^2-4)|-ln(2)
Since ln(2) is constant, so we can omit it and add a new constant C to the solution,
=ln|x+sqrt(x^2-4)|+C