是：∫（1+2^t)^0.5dt
令1+2^t=x
t=log2(x-1)
dt=1/[(x-1)ln2]dx
原式=1/ln2*∫x^0.5/(x-1)dx
=2/3*1/ln2*∫1/(x-1)d(x^3/2)令：x^3/2=u.
=2/(3ln2)*∫1/(u^1/3+1)(u^1/3-1)du
=1/3ln2*[∫1/(u^1/3-1)du-∫(u^1/3+1)du]令：u^1/3=q
=1/ln2*[∫q^2/(q-1)dq-∫q^2(q+1)dq]
=1/ln2*∫q^2/(q^2-1)dq
=1/ln2*[∫1dq+∫1/(q^2-1)dq]
下面我不做了,你自己可以解出来.