高数不定积分∫1/(x+1)√xdx

Question

如题所述

Answer 1

∫1/(x+1)√xdx中√x在分母吗。若是
∫1/[(x+1)√x]
dx=2∫1/(x+1)d√x=2∫1/[(√x)^2+1]d√x=2arc
tan√x+C
若是
∫1/(x+1)*√xdx=∫x/(x+1)*1/√xdx=2∫x/(x+1)d√x=2∫[1-1/(x+1)]d√x
=2√x-2arc
tan√x+C
这种积分通常换元为根式来作，供你参考