求不定积分∫(1+x^2)^(-1/2)dx

∫(1+x²)^(-1/2)dx
=∫1/(1+x²)^(1/2)dx
=∫1/√(1+x²)dx,
令x=tanU,则dx=sec²UdU
=∫[1/√(1+tan²U)•sec²U]dU
=∫sec²U/√(sec²U) dU
=∫secUdU
=ln|secU+tanU|+C
由于设了tanU=x,
根据直角三角形勾股定理：对边是x,邻边是1,斜边是√(1+x²)
那么secU=r/x=√(1+x²)/1=√(1+x²).
原积分=ln|x+√(1+x²)|+C
附：∫secUdU=ln|secU+tanU|+C推导过程
∫secxdx
=∫sec²x/secxdx
=∫cosx/cos²xdx
=∫1/cos²xdsinx
=∫1/(1-sin²x)dsinx
=-∫1/(sinx+1)(sinx-1)dsinx
=-∫[1/(sinx-1)-1/(sinx+1)]/2dsinx
=-[∫1/(sinx-1)dsinx-∫1/(sinx+1)dsinx]/2
=[∫1/(sinx+1)d(sinx+1)-∫1/(sinx-1)d(sinx-1)]/2
=(ln|sinx+1|-ln|sinx-1|)/2+C
=ln√|(sinx+1)/(sinx-1)|+C
=ln√|(sinx+1)²/(sinx+1)(sinx-1)|+C
=ln√|(sinx+1)²/(sin²x-1)|+C
=ln√|-(sinx+1)²/cos²x|+C
=ln|(sinx+1)/cosx|+C
=ln|tanx+1/cosx|+C
=ln|secx+tanx|+C