Let x = tante, t∈(-π/2, π/2)

√( 1+x？ )=sect，dx=sec？ terminal transferase

√( 1+x？ )dx

= ∫ seconds? t dt

=∫d section (tant)?

=sect*tant-∫tant d(sect)？

=sect*tant-∫tan？ t * sectdt？

= sect * tante-∫ (seconds? t- 1)* secdt？

=sect*tant-∫sec？ tdt+∫sectdt？

∫sec^3tdt=( 1/2)(sect*tant+∫sectdt)

=( 1/2)(Sect * tant+ln | Sect+tant |)+C？

The original formula = (1/2) [x * √ (x 2+1)+ln | √ (x 2+1)+x |]+C.

The meaning of indefinite integral:

A function can have indefinite integral without definite integral or definite integral without definite integral. A continuous function must have definite integral and indefinite integral.

If there are only finite discontinuous points on the finite interval [a, b] and the function is bounded, then the definite integral exists; If there are jumping points, going points and infinite discontinuous points, the original function must not exist, that is, the indefinite integral must not exist.