Question

数学求不定积分积分？

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Answer 1

Answer 2

(1)
let
u= e^(x/2)
du = (1/2)e^(x/2) dx
dx = 2du/u
let
1/[u^2.(1 + u)] ≡ A/u +B/u^2 +C/(1+u)
=>
1 ≡ Au(1+u) +B(1+u) +Cu^2
u=0, => B= 1
u=-1, => C =1
coef. of u
A+B=0
A+1=0
A=-1
1/[u^2.(1 + u)]
≡ A/u +B/u^2 +C/(1+u)
≡ -1/u +1/u^2 +1/(1+u)
∫ dx/ [ e^(x/2) + e^x]
=∫ [2du/u] / (u + u^2)
=2∫ du / [u^2.(1 + u)]
=2∫ [-1/u +1/u^2 +1/(1+u) ] du
=2[ ln|(1+u)/u| - 1/u ] +C
=2[ ln|(1+e^(x/2))/e^(x/2)| - 1/e^(x/2) ] +C
=2[ ln|1+e^(x/2)| -(1/2)x - e^(-x/2) ] +C
(2)
∫x^2.arctanx/(1+x^2) dx
=∫[ arctanx - arctanx/(1+x^2) ] dx
=∫ arctanx dx - (1/2)(arctanx)^2
=xarctanx - ∫ x/(1+x^2) dx - (1/2)(arctanx)^2
=xarctanx - (1/2)ln|1+x^2| - (1/2)(arctanx)^2 + C