Question

高数:(1)-(2)怎么做，求不定积分需要详细过程，急急

Answer 1

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

let
2x/[(x+1)^2(x-1)]≡ A/(x-1) + B/(x+1) + C/(x+1)^2
2x≡ A(x+1)^2 + B(x-1)(x+1) + C(x-1)

x=-1, C= 1
x=1, A=1/2

coef. of x^2
A+B=0
B=-1/2

2x/[(x+1)^2(x-1)]≡ (1/2)[1/(x-1)] -(1/2)[ 1/(x+1)] + 1/(x+1)^2

∫(x^2+1)/[(x+1)^2(x-1)] dx
=ln|x-1| -∫2x/[(x+1)^2(x-1)] dx
=ln|x-1| -( (1/2)ln|x-1| -(1/2)ln|x+1| + arctanx ) + C
=(1/2)ln|(x-1)(x+1) | - arctanx + C

(2)
x^3+1 = (x+1)(x^2-x+1)
let

1/(x^3+1)≡ A/(x+1) + (Bx+C)/(x^2-x+1)
1≡ A(x^2-x+1)+(Bx+C)(x+1)

x=-1, A=1/3

coef. of x^2

A+B=0
B =-1/3

coef. of constant
A+C=1
C= 2/3

∫3/(x^3+1) dx
=∫ [1/(x+1) - (x-2)/(x^2-x+1) ]dx
=ln|x+1| - (1/2)∫(2x-1)/(x^2-x+1) dx + (3/2)∫ dx/(x^2-x+1)
=ln|x+1| - (1/2)ln|x^2-x+1|+ (3/2)∫ dx/(x^2-x+1)

consider
x^2-x+1 = 3/4 +(x-1/2)^2
let
x+1/2 =(√3/2)tany
dx =(√3/2)(secy)^2 dy

∫ dx/(x^2-x+1)
=(2√3/3)∫ dy
=(2√3/3)y + C'
=(2√3/3)arctan[(2x+1)/√3] + C'

∫3/(x^3+1) dx
=ln|x+1| - (1/2)ln|x^2-x+1|+ (3/2)∫ dx/(x^2-x+1)
=ln|x+1| - (1/2)ln|x^2-x+1|+ √3arctan[(2x+1)/√3] + C

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