Question

求不定积分 1／（1＋2x）（1＋x^2）dx

答案是5／2ln|1＋2x|－1／5（1＋x^2）＋1／5arctanx＋C
具体过程怎么写

Answer 1

设1/[(1+2x)(1+x^2)]
=A/(1+2x)+(Bx+C)/(1+x^2)
=[A(1+x^2)+(Bx+C)(1+2x)]/[(1+2x)(1+x^2)]
=(A+Ax^2+Bx+2Bx^2+C+2Cx)/[(1+2x)(1+x^2)]
=[(A+2B)x^2+(B+2C)x+A+C]/[(1+2x)(1+x^2)]
对应系数相等，得：
A+2B=0，B+2C=0，A+C=1
解得A=4/5，B=-2/5，C=1/5
则1/[(1+2x)(1+x^2)]=(4/5)/(1+2x)+[(-2/5)x+(1/5)]/(1+x^2)
∫1/[(1+2x)(1+x^2)]dx
=∫(4/5)/(1+2x)dx+∫(-2/5)x/(1+x^2)dx+∫(1/5)/(1+x^2)dx
=(2/5)∫1/(1+2x)d(1+2x)-(1/5)∫1/(1+x^2)d(1+x^2)+(1/5)∫1/(1+x^2)dx
=(2/5)ln|1+2x|-(1/5)ln(1+x^2)+(1/5)arctanx+C

Answer 2

令1/[(1 + 2x)(1 + x^2)] = A/(1 + 2x) + (Bx + C)/(1 + x^2)
则1 = A(1 + x^2) + (Bx + C)(1 + 2x)
1 = (A + 2B)x^2 + (B + 2C)x + (A + C)
A + 2B = 0、B + 2C = 0、A + C = 1
解得A = 4/5、B = - 2/5、C = 1/5
原式 = (4/5)∫ dx/(1 + 2x) - (2/5)∫ x/(1 + x^2) dx + (1/5)∫ dx/(1 + x^2)
= (4/5)(1/2)∫ d(1 + 2x)/(1 + 2x) - (2/5)(1/2)∫ d(1 + x^2)/(1 + x^2) + (1/5)∫ dx/(1 + x^2)
= (2/5)ln|1 + 2x| - (1/5)ln(1 + x^2) + (1/5)arctan(x) + C