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

Question

10033 views around the world

You can reuse this answer
Creative Commons License

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

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