Question

1/(X2+3)的反导数十啥?（x平方+3）分之1 的反导

Answer 1

1/(x²+3)的反导数即求1/(x²+3)的不定积分

∫[1/(x²+3)]dx,x=√3tanu,dx=√3sec²udu

=√3∫{(sec²u)/[(√3tanu)²+3]}du

=√3∫[(sec²u)/(3tan²u+3)]du

=(√3/3)∫(sec²u/sec²u)du

=(1/√3)∫du

=(1/√3)u+C,∵x=√3tanu∴u=arctan(x/√3)

=(1/√3)arctan(x/√3)+C