Original formula = ∫1[secu * tanu] * sekutanudu = ∫1du = u+c = arccos (1/x)+C.

When x; 1，dx=-dt

∫ 1/[x √( x ^ 2- 1)]dx =∫ 1/[(-t)√( t ^ 2- 1)]d(-t)=∫。

Two combinations are obtained: arccos1| x |+c.

2. Similarly, when x> is 1, let x=secu, then (1/x)=cosu, then √ (x 2- 1) = TANU, dx=secutanudu.

The original formula = ∫ (tanu/secu) * secutanu du = ∫ (tanuj) 2du = ∫ [(secuj) 2-1] du = tanu-u+c.

=√(x^2- 1)-arccos( 1/x)+c

When x; 1，dx=-dt

∫√(x^2- 1)/x dx =∫√(t^2- 1)/(-t)d(-t)=∫√(t^2- 1)/t dt

Exactly the same as just now. By directly using the results just now, we get: √ (t2-1)-arccos (1/t)+c = √ (x2-1)+arccos (1/x)+c.

Your answer to this question is wrong.