(基本)積分問題一條

設w=cos(2π/5)+i sin(2π/5), p=2π/5, q=4π/5, 則w^5=1 且
x^5+1=(x+1)(x+w)(x+w^2)(x+w^3)(x+w^4)
5/(x^5+1)=(partial fraction)
= 1/(x+1) + w/(x+w) + w^4/(x+w^4) + w^2/(x+w^2) + w^3/(x+w^3)
=1/(x+1)+ (2xcosp+2)/(x^2+2xcosp+1) + (2xcosq+2)/(x^2+2xcosq+1)
=1/(x+1)+2[cosp(x+cosp)+(sinp)^2]/[(x+cosp)^2+(sinp)^2]
+2[cosq(x+cosq)+(sinq)^2]/[(x+cosq)^2+(sinq)^2]
故
∫ 5/(x^5+1) dx=ln|x+1|+cosp ln(x^2+2xcosp+1)+ 2sinp arctan[(x+cosp)/sinp]
+cosq ln(x^2+2xcosq+1)+2sinq arctan[(x+cosq)/sinq]+C
∫ dx/(x^5+1)
=(1/5)ln|x+1|+(1/5)cosp ln(x^2+2xcosq+1)+(2/5)sinp arctan[(x+cosp)/sinp]
+(1/5)cosq ln(x^2+2xcosq+1)+(2/5)sinq arctan[(x+cosq)/sinq]+C

很強很暴力

　　　　

答案很恐怖><"
http://integrals.wolfram.com/index.jsp?expr=1%2F%28x%5E5%2B1%29&random=false