Question

高等数学积分问题，第一步是怎么化成第二步的？谢谢

Answer 1

令x=tant,计算化简，再积分，积分完再带入t=arctanx

Answer 2

1= (1/2) [ (x^2+1) -(x^2-1) ]
∫ dx/(1+x^4)
=(1/2)[ ∫ (x^2+1)/(1+x^4)dx -∫(x^2-1)/(1+x^4) dx ]

追问

这一步然后，是怎么化成最后的积分的呢？

追答

let
tanu =( x- 1/x) /√2
√2 (secu)^2 du = ( 1 + 1/x^2) dx
∫ (x^2+1)/(1+x^4)dx
=∫ [(x^2+1)/x^2 ]/[ (1+x^4)/x^2] dx
=∫ [1+1/x^2 ]/[ (x^2 + 1/x^2) ] dx
=∫ [1+1/x^2 ]/[ (x- 1/x)^2+2 ] dx
= ∫ √2 (secu)^2 du /[ 2(tanu)^2 +2 ]
=√2/2 )∫ (secu)^2 du/[ (tanu)^2 +1 ]
=(√2/2) ∫ d(tanu)/[ 1+(tanu)^2 ]
=(√2/2) u
=(√2/2) arctan[( x- 1/x) /√2] + C1
/
let
v = x+ 1/x
dv = (1 -1/x^2) dx
∫(x^2-1)/(1+x^4) dx
=∫(1- 1/x^2)/( x^2 +1/x^2 ) dx
=∫(1- 1/x^2)/[ ( x +1/x )^2 -2 ] dx
=∫dv/(v^2 -2)
=[1/(2√2)]∫[1/(v -√2) - 1/(v +√2) ] dv
=[1/(2√2)] ln|(v -√2)/(v +√2)| +C2
=[1/(2√2)] ln|( x+ 1/x -√2)/( x+ 1/x +√2)| +C2

Answer 3

忘了，去年还考八十分呢，一个暑假就蒙圈了