Question

换元积分法求不定积分

Answer 1

Answer 2

令 根号x=t ,则x=t²,dx=2tdt
所以原式= ∫2t/(1+t) dt
=2∫(t+1-1)/(t+1) dt
=2 ( ∫dt - 1/(t+1)dt )
=2(t - ln|t+1|) + C
有原式= 2根号x- 2ln(根号x+1) +C