Question

用换元法计算积分1/[根号（x+1)]乘以（2+x）

Answer 1

令t=√(x+1)
则x=t^2-1,dx=2tdt
∫1/[(x+2)*√(1+x)]dx
=∫[1/(t^2+1)*t]*(2tdt)
=∫2/(1+t^2)dt
=2∫1/(1+t^2)dt
=2arctant+C
=2arctan√(x+1)+C