Question

求定积分∫1/（x^4+1)dx

Answer 1

∫dx/(x^4+1)=∫dx/[(x^2+1)^2-2x^2]
=∫dx/[(x^2+1-√2x)(x^2+1+√2x)]
=∫(1/2√2x)[ (x^2+1+√2x)-(x^2+1-√2x)]dx/[(x^2+1+√2x)(x^2+1-√2x)]
=∫(1/(2√2x))dx/(x^2+1-√2x) - ∫(1/(2√2x))dx/(x^2+1+√2x)
=(1/(2√2))[∫(1/2)d(x^2+1-√2x)/(x^2+1-√2x) +∫(1/2)*√2dx/(x^2+1-√2x)
-∫(1/2)d(x^2+1+√2x)/(x^2+1+√2x)+∫(1/2)*√2dx/(x^2+1+√2x)]
=(1/(2√2))*[(1/2)(ln|(x^2+1-√2x)|-ln|(x^2+1+√2x)|)
+(1/4)∫dx/[(x-√2/2)^2+1/2] +(1/4)∫dx/[(x+√2/2)^2+1/2]]
=(√2/4)[(1/2)(ln|(x^2+1-√2x)| -ln|(x^2+1+√2x)|)
+(√2/4)arctan(√2x-1)
+(√2/4)arctan(√2x+1)]+C
=(√2/8)[ln|(x^2+1-√2x)| -ln|(x^2+1+√2x)|)]
+(1/8)arctan(√2x-1)
+(1/8)arctan(√2x+1)+C
参见http://zhidao.baidu.com/question/305672529.html

Answer 2

(⊙o⊙)…