Question

第二题不定积分

Answer 1

解：换元法
令u=(e^x+1)^1/2
u^2=e^x+1
e^x=u^2-1
x=ln(u^2-1）
dx=1/(u^2-1)x2u=2u/(u^2-1)du
原是=积分ux2u/(u^2-1)du
=2积分u^2/(u^2-1)du
=2积分（u^2-1+1)/(u^2-1)du
=2积分（1+1/(u^2-1))du
=2（积分1du+积分1/(u^2-1)du)
=2(u+积分1/(u^2-1)du)
=2u+2积分1/(u^2-1)du
积分1/(U^2-1)du
=1/2积分（1/(u-1)-1/(u+1))du
=1/2（积分1/(u-1)du-积分1/(u+1)du)
=1/2(ln/u-1/-ln/u+1/)+C
=1/2ln/(u-1)/(u+1)/+C
原是=2u+2x1/2ln/(u-1)/(u+1)/+C
=2u+ln/(u-1)/(u+1)/+C.
u=（e^x+1)^1/2
=2（e^x+1)^1/2+ln/(（e^x+1)^1/2-1)/(（e^x+1)^1/2+1)/+C.
答：答案是2（e^x+1)^1/2+ln/(（e^x+1)^1/2-1)/(（e^x+1)^1/2+1)/+C.