Complete the square for #x^2-2x-8#:
#(x^2-2x)-8=0#
#(x^2-2x)=8#
#(x^2-2x+(2/2)^2)=8+(2/2)^2#
#(x^2-2x+1)=8+1#
#(x^2-2x+1)=9#
#(x-1)^2-9=0#
Rewrite the integral using the completed square:
#int1/sqrt((x-1)^2-9)dx#
#u=x-1#
#du=dx#
Rewrite the integral using the above substitution:
#int1/sqrt(u^2-9)du#
We can use trig substitution here:
#u=3sec(theta)#
#du=3sec(theta)tan(theta)d theta#
Rewrite the integral using the trig substitution:
#int(3sec(theta)tan(theta))/sqrt(9(sec^2(theta)-1))d theta#
#int(cancel3sec(theta)tan(theta))/(cancel3sqrt(sec^2(theta)-1))d theta#
#int(sec(theta)tan(theta))/(sqrt(sec^2(theta)-1))d theta#
Recall the identity:
#1+tan^2(theta)=sec^2(theta)#
Therefore:
#sec^2(theta)-1=tan^2(theta)#
Rewrite the integral with the identity:
#int(sec(theta)tan(theta))/(sqrt(tan^2(theta)))d theta#
#int(sec(theta)canceltan(theta))/(canceltan(theta))d theta#
#int(sec(theta)d theta#
#ln|sec(theta)+tan(theta)|+C#
#u=3sec(theta)#, so #sec(theta)=u/3#
If #sec(theta)=u/3#, #sec^2(theta)=u^2/9#
Using the aforementioned identity:
#tan^2(theta)=sec^2(theta)-1#
#tan^2(theta)=u^2/9-1#
#tan(theta)=sqrt(u^2/9-1)#
Rewrite our integral in terms of #u:#
#ln|u/3+sqrt((u^2-9)/9)|+C#
#ln|u/3+sqrt((u^2-9))/3|+C#
#ln|u+sqrt(u^2-9)|-ln|3|+C# using logarithmic rules for division.
#ln|3|=ln(3)#, and this just becomes part of our constant #C# because it's a constant value.
#ln|u+sqrt(u^2-9)|+C#
Rewrite in terms of #x:#
#ln|x-1+sqrt((x-1)^2-9)|+C#
