Question

求定积分:[(2+根号下x)分之1]dx,上限是1,下限是0?

Answer 1

令根号x=t

x=t²,dx=2tdt

x=0,t=0;x=1,t=1

所以

原式=∫(0,1)1/(2+t) *2tdt

=2∫（0,1）t/(2+t)dt

=2∫（0,1）(t+2-2)/(2+t)dt

=2∫（0,1）(1-2/(t+2))dt

=(2t-4ln|t+2|)|(0,1)

=2-4ln3-(0-4ln2)

=2-4ln(3/2)