According to the equivalent infinitesimal:

√( 1+x) - 1 ~x/2

Primitive restriction

=lim(x→0) ∫(sin？ x，0) ln( 1+t)dt / [(x^4)/2]

Look up the function y=ln( 1+t) in [0, sin? The integral on x] obviously satisfies the integral mean value theorem, so:

∫ (sin? x，0) ln( 1+t)

=ln( 1+ε) (sin？ x-0)

= sin? xln( 1+ε)

Where: ε∈[0, sin? x]

Therefore:

Primitive restriction

=lim(x→0) sin？ xln( 1+ε) / [(x^4)/2]

According to the equivalent infinitesimal:

sinx ~x

ln( 1+x) ~x

When x→0, the interval [0, sin? X] approaches 0, ε and sin? X is also close to zero, so:

ε ~ sin? x

So:

Primitive restriction

=lim(x→0) x？ ε/[(x^4)/2]

=lim(x→0) x？ x？ /[(x^4)/2]

=2

This problem can also be solved by Robida's law, and molecules can be derived by variable limit integral!

Primitive restriction

=lim(x→0) ln( 1+sin？ x) 2sinxcosx / 2x？

=lim(x→0) x？ 2 2x// 2x？

=2