Question

求f(x)=1/(1+根号下x)的不定积分

Answer 1

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

Answer 2

由题意，可设√x=t,t²=x,dx=2tdt (t>=0)
所以∫f(x)=∫dx/(1+√x)
=∫2tdt/(1+t)
=2∫(t+1-1)dt/(t+1)
=2[∫dt-∫dt/(t+1)]
=2[t-ln(t+1)]+C=2[√x-ln(√x+1)]+C

Answer 3

令√x=t,
x=t^2,dx=2tdt,
f(x)=∫2t/(1+t^2)dt
=ln(1+t^2)
=ln(1+x)

Answer 4

换元法：令t=根号x,那么x=t^2,dx=d(t^2)=2tdt,
那么原式=积分[2t/(1+t)]dt=积分[(2(t+1)-2)/(1+t)]dt
=积分2dt-2积分[1/(1+t)]dt=2t-2ln(t+1)+C=2倍根号x-2ln(1+根号下x)+C