过程很简单用第二类换元积分法便可解决请看图:
原式=∫[x/√(x^4+1)]dx+∫{1/[x√(x^4+1)]}dx
原式=(1/2)∫[1/√(u^2+1)]du+(1/2)∫{1/[u√(u^2+1)]}du
=(1/2)∫[cosθ/(cosθ)^2]dθ+(1/4)∫{(t+1-t+1)/[(t+1)(t-1)]}dt
-(1/4)∫[1/(t+1)]dt
+(1/4)ln|t-1|-(1/4)ln|t+1|
+(1/4)ln|t-1|-(1/4)ln|t+1|
+(1/4)ln|√(u^2+1)-1|-(1/4)ln|√(u^2+1)+1|+C
+(1/4)ln|√(x^4+1)-1|-(1/4)ln|√(x^4+1)+1|+C
+(1/4)ln|√(x^4+1)-1|-(1/4)ln|√(x^4+1)+1|+C