求不定积分dx／x(x＾2+1)要详细过程谢了答案为lnx/根号下1+x＾2+C

解
设x=tant,则dx=sect²*dt,t=arctanx
∫dx／x(x＾2+1)
=∫sec²t/tantsec²t dt
=∫1/tantdt
=∫cost/sintdt
=∫1/sintdsint
=ln │sint│+C
=ln√（1- cos² t）+C
=ln√（1-1/sec²t）+C
=ln√[1-1/（tan²t+1）]+C
=ln√[1-1/（1+x＾2）+C
=ln√x＾2/（1+x＾2）+C
=lnx/根号下1+x＾2+C

∫dx/[x(x＾2+1)]
let
1/[x(x＾2+1)] = A/x + (Bx+C)/(x＾2+1)
=> 1 = A(x^2+1) + (Bx+C)x
put x=0
A=1
coef. of x
C=0
coef. of x^2
A+B=0
B=-1
=>
1/[x(x＾2+1)] = 1...