x^4+4=(x^2+2)^2-4x^2=(x^2+2x+2)(x^2-2x+2)
なので
1/(x^4+4)=(1/8){(x+2)/(x^2+2x+2)}-(1/8){(x-2)/(x^2-2x+2)}
(x^2)/(x^4+4)=(1/4){x/(x^2-2x+2)}-(1/4){x/(x^2+2x+2)}
と部分分数分解できます。
[前半の設問]
Ia=∫1/(x^4+4)dx
=(1/8){∫(x+2)/(x^2+2x+2)dx-∫(x-2)/(x^2-2x+2)dx}
=I1-I2
I1=(1/8)∫{(x+2)/(x^2+2x+2)dx
x^2+2x+2=(x+1)^2+1,x+1=tで置換積分して
(途中略)
=(1/16)log(x^2+2x+2)+(1/8)arctan(x+1)+c1
I2=(1/8)∫{(x-2)/(x^2-2x+2)dx
x^2-2x+2=(x-1)^2+1,x-1=tで置換積分して
(途中略)
=(1/16)log(x^2-2x+2)-(1/8)arctan(x-1)+c2
∴Ia=I1-I2
=(1/16)log{(x^2+2x+2)/(x^2-2x+2)}
+(1/8){arctan(x+1)+arctan(x-1)}+C
=(1/8)log{(x^2+2x+2)/(x^4+4)}+(1/8){arctan(x+1)+arctan(x-1)}+C
(C=c1-c2:積分定数)
[後半の設問]
Ib=∫x^2/(x^4+4)dx
=(1/4){∫x/(x^2-2x+2) dx-∫{x/(x^2+2x+2)}dx}
=I3-I4
I3=(1/4)∫x/(x^2-2x+2) dx
x^2-2x+2=(x-1)^2+1,x-1=tで置換積分して
(途中略)
=(1/8)log(x^2-2x+2)+(1/4)arctan(x-1)+c3
I4=(1/4)∫x/(x^2+2x+2) dx
x^2+2x+2=(x+1)^2+1,x+1=tで置換積分して
(途中略)
=(1/8)log(x^2+2x+2)-(1/4)arctan(x+1)+c4
∴Ib=I3-I4
=(1/8)log{(x^2-2x+2)/(x^2+2x+2)}
+(1/4){arctan(x-1)+arctan(x+1)} +C
=(1/4)log(x^2-√2x+2)-(1/8)log(x^4+4)
+(√6/12){arctan(x-1)+arctan(x+1)} +C
(C=c3-c4:積分定数)
お礼
解けました!ありがとうございます(>_<)