Question

请问不定积分∫dm /(1+m∧4)怎么求？

Answer 1

分母填两项2x²和-2x²以配成平方差：
∫ [1/ (x⁴+ 2x² +1 - 2x²)] dx =
∫ {1/ [(x⁴+ 2x² +1) - 2x²]} dx =
∫ {1/ [(x² +1)² - 2x²]} dx =
对分母因式分解得：
∫ {1/ [(x² +1)² - (√2x)²]} dx =
∫ {1/ {[(x² +1) + (√2x)] [(x² +1) - (√2x)]}} dx =
∫ {1/ [(x² + √2x +1)(x² - √2x +1)]} dx =
现在可以用部分分式积分法了：
1/ [(x² +√2x +1)(x² -√2x +1)] = (Ax + B) /(x² +√2x +1) + (Cx + D) /(x² -√2x +1) →
1/ [(x² +√2x +1)(x² -√2x +1)] =
[(Ax + B)(x² -√2x +1) + (Cx + D)(x² +√2x +1)] / [(x² +√2x +1)(x² -√2x +1)] →
整理分子：
1 = [(Ax + B)(x² -√2x +1) + (Cx + D)(x² +√2x +1)] →
1 = Ax³ -√2Ax² + Ax + Bx² -√2Bx + B + Cx³+ √2Cx²+ Cx + Dx² +√2Dx + D →
1 = (A + C)x³ + (-√2A + B + √2C + D)x²+ (A -√2B + C +√2D)x + (B + D) →
解以下方程组：
| A + C = 0 → A = - C → A = 1/(2√2)
| -√2A + B +√2C+ D = 0 → -√2(- C)+ B +√2C+ D = 0 → 2√2C + 1 = 0 → C = -1/(2√2)
| A -√2B + C +√2D = 0 → - C -√2B + C +√2D = 0 → √2D = √2B → D = B → B = (1/2)
| B + D = 1 → D + D = 1 → D = (1/2)
方程组的解代入各个分子得：
1/ [(x² +√2x +1)(x² -√2x +1)] = (Ax + B) /(x² +√2x +1) + (Cx + D) /(x² -√2x +1) →
1/ [(x² +√2x +1)(x² -√2x +1)] =
{[1/(2√2)]x + (1/2)} /(x² +√2x +1) + {[-1/(2√2)]x + (1/2)} /(x² -√2x +1) →
1/ [(x² +√2x +1)(x² -√2x +1)] =
[1/(2√2)] [(x +√2) /(x² +√2x +1)] + [1/(2√2)] [(-x +√2) /(x² -√2x +1)] →
所以∫ {1/ [(x² + √2x +1)(x² - √2x +1)]} dx =
∫ {[1/(2√2)] [(x +√2) /(x² +√2x +1)] + [1/(2√2)] [(-x +√2) /(x² -√2x +1)]} dx =
[1/(2√2)] ∫ [(x +√2) /(x² +√2x +1)] dx + [1/(2√2)] ∫ [(-x +√2)} /(x² -√2x +1)] dx =

[1/(2√2)](1/2) ∫ [2(x+√2) /(x²+√2x+1)] dx+ [1/(2√2)](-1/2) ∫ [(-2)(-x+√2)} /(x²-√2x+1)] dx =

[1/(4√2)] ∫ [(2x + 2√2) /(x²+√2x +1)] dx - [1/(4√2)] ∫ [(2x - 2√2)} /(x²-√2x +1)] dx =

[1/(4√2)] ∫ [(2x +√2 +√2) /(x²+√2x +1)] dx - [1/(4√2)] ∫ [(2x - √2 - √2)} /(x²-√2x +1)] dx =

[1/(4√2)] ∫ {[(2x +√2)/(x²+√2x +1)] + [√2 /(x²+√2x +1)]} dx -
[1/(4√2)] ∫ {[(2x - √2)/(x²-√2x +1)] - [√2 /(x²-√2x +1)]} dx =

[1/(4√2)] ∫ [(2x +√2)/(x²+√2x +1)] dx + [1/(4√2)] ∫ [√2 /(x²+√2x +1)] dx -
[1/(4√2)] ∫ [(2x - √2)/(x²-√2x +1)] dx + [1/(4√2)] ∫ [√2 /(x²-√2x +1)] dx =

[1/(4√2)] ∫ [d(x²+√2x +1)] /(x²+√2x +1) + (1/4) ∫ [1 /(x²+√2x +1)] dx -
[1/(4√2)] ∫ [d(x²-√2x +1)] /(x²-√2x +1)] + (1/4) ∫ [1 /(x²-√2x +1)] dx =

[1/(4√2)] ln (x²+√2x +1) + (1/4) ∫ [1 /(x²+√2x +1)] dx - [1/(4√2)] ln (x²-√2x +1) +
(1/4) ∫ [1 /(x²-√2x +1)] dx =

[1/(4√2)] [ln (x²+√2x +1) - ln (x²-√2x +1)] + (1/4) ∫ [1 /(x²+√2x +1)] dx +
(1/4) ∫ [1 /(x²-√2x +1)] dx =
根据对数性质，
[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + (1/4) ∫ [1/(x²+√2x +1)] dx +
(1/4) ∫ [1/(x²-√2x +1)] dx =

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + (1/4)(2) ∫ dx/[2(x²+√2x +1)] +
(1/4)(2) ∫ dx/ [2(x²-√2x +1)] =

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + (1/2) ∫ dx/(2x²+2√2x +2) +
(1/2) ∫ dx/ (2x²-2√2x +2) =

2写成1+1是为了配方，
[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + (1/2) ∫ dx/(2x² + 2√2x +1 +1) +
(1/2) ∫ dx/ (2x²- 2√2x +1 + 1) =

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + (1/2) ∫ dx/[(2x² + 2√2x +1) +1] +
(1/2) ∫ dx/ [(2x²- 2√2x +1) +1] =

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + (1/2) ∫ dx/[(√2x +1)²+1] +
(1/2) ∫ dx/ [(√2x -1)²+1] =

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + (1/2)(1/√2) ∫ (√2dx)/[(√2x +1)² +1] +
(1/2)(1/√2) ∫ (√2dx)/ [(√2x -1)² +1] =

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + [1/(2√2)] ∫ [d(√2x +1)] /[(√2x +1)² +1] +
[1/(2√2)] ∫ [d(√2x - 1)] / [(√2x -1)² +1] =

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + [1/(2√2)] arctan (√2x +1) +
[1/(2√2)] arctan (√2x - 1) + c

1/(x⁴+ 1) dx 的不定积分就是 [1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] +
[1/(2√2)] arctan (√2x +1) + [1/(2√2)] arctan (√2x - 1) + c

Answer 2

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