How do you find the integral of #dx/sqrt(x^2-4)#?

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Answer 1 · 2018-03-15T09:13:21

#ln|((x/2)+sqrt((x^2/2-1)))|+c#

#int1/(sqrt(x^2-4))dx#

we can either use a trig substitution

using #tan^2theta=sec^2theta-1#

#x=2secu=>dx=2secutanudu#

#int1/(sqrt(x^2-4))dx=int1/((sqrt(4sec^2u-4)))(2secutanudu)#

#=int(cancel(2)secucancel(tanu))/(cancel(2sqrt(sec^2u-1)))du#

#=intsecudu#

this isa standard integral

#=ln|(secu+tanu)|+c#

substituting back

#=ln|((x/2)+sqrt((x^2/2-1)))|+c#

which can be simplified as required