Question

求（1+x^2)/(1+x^4)的在（-1,1）上的定积分

Answer 1

(1+x^2)/(1+x^4)在(-1,1)上的定积分
∫(1+x^2)/(1+x^4)dx
=∫(x^2+1)/[(x^2+√2x+1)(x^2-√2x+1)]dx
设(x^2+1)/[(x^2+√2x+1)(x^2-√2x+1)]
=(Ax+B)/(x^2+√2x+1)+(Cx+D)/(x^2-√2x+1)
待定系数法，解得
A=C=0
B=D=1/2
所以有
(x^2+1)/[(x^2+√2x+1)(x^2-√2x+1)]
=1/2(x^2+√2x+1)+1/2(x^2-√2x+1)
原不定积分等价于
1/2*[∫1/(x^2+√2x+1)dx+∫1/(x^2-√2x+1)dx]
=∫1/[(√2x+1)^2+1]dx+∫1/[(√2x-1)^2+1]dx
=√2/2*∫1/[(√2x+1)^2+1]d(√2x)+√2/2*∫1/[(√2x-1)^2+1]d(√2x)
=√2/2*[arctan(√2x+1)+arctan(√2x-1)]
又有原被积函数为偶，积分区间对称
所以定积分等于
√2[arctan(√2+1)+arctan(√2-1)]
=√2(3π/8+π/8)
=√2π/2
手机纯手打，真的不容易啊。
希望对楼主有所帮助，望采纳