Question

求∫dx/1+(根号x)

求高手

Answer 1

解
令√x=t
则t²=x，dx=2tdt
∴∫dx/(1+√x)
=∫2tdt/(t+1)
=2∫[(t+1)-1]/(t+1)dt
=2∫1-1/(t+1)dt
=2t-2ln|t+1|+C
=2√x-2ln(√x+1)+C

追问

=2∫1-1/(t+1)dt 不理解

追答

=2∫[1-1/(t+1)]dt——1减去(t+1分之1)