Question

不定积分∫dx/((2x-1)^{1/2}+1)=____

令(2x-1)^{1/2}=t，则x=(t^2+1)/2，dx=tdt
∫dx/((2x-1)^{1/2}+1)
=∫tdt/(t+1)
令t+1=u，则t=u-1，dt=du
∫tdt/(t+1)
=∫(u-1)du/u
=∫du - ∫du/u
=u - ln|u| + C
=(2x-1)^{1/2}+1 - ln|(2x-1)^{1/2}+1| + C

哪里错了？

Answer 1

x=tant,dx=sec²tdt
∫dx/[(2x^2+1)(x^2+1)^(1/2) ]
=∫sec²tdt/[(2tan²t+1)sect]
=∫dt/[cost((2sin²t/cos²t)+1)]
=∫costdt/[((2sin²t+cost²)]
=∫[1/(1+sin²t)]d(sint)
=arctan(sint)+C
三角替换有sint=x/√(1+x²)
所以原不定积分
∫dx/(2x^2+1)(x^2+1)^(1/2)
=arctan[x/√(1+x²)]+C

Answer 2

涉江采芙蓉,兰泽多芳草.采之欲遗谁?所思在远道.