分母を因数分解して部分分数分解するといいでしょう。
1/(x^4+1)=1/((x^2+1)^2-2x^2)=1/(x^2+1+√2x)(x^2+1-√2x)
=(√2/4){(x+√2)/(x^2+1+√2x)-(x-√2)/(x^2+1-√2x)}
=(√2/4){(x+1/√2+1/√2)/(x^2+1+√2x)-(x-1/√2-1/√2)/(x^2+1-√2x)}
x+1/√2=u, x-1/√2=v とおくと
=(√2/4){(u+1/√2)/(u^2+1/2)-(v-1/√2)/(v^2+1/2))}
であるから
I=∫1/(x^4+1)dx
=(√2/4) { ∫(u+1/√2)/(u^2+1/2) du-∫(v-1/√2)/(v^2+1/2)dv}
=(√2/4) {(1/2)ln(u^2+1/2)+tan^-1(√2u) -(1/2)ln(v^2+1/2)+tan^-1(√2v)}+C
=(√2/8) ln{(u^2+1/2)/(v^2+1/2)}+(√2/4) {tan^-1(√2u)+tan^-1(√2v)}+C
=(√2/8) ln{(u^2+1/2)/(v^2+1/2)}+(√2/4) tan^-1{(√2(u+v))/(1-2uv)}+C
u,v を x に戻して
I=(√2/8) ln{(x^2+√2 x+1)/(x^2-√2 x+1)}+(√2/4) {tan^-1(2√2x/(1-2(x^2-1/2))}+C
=(√2/8) ln{(x^2+√2 x+1)/(x^2-√2 x+1)}+(√2/4) tan^-1(√2x/(1-x^2))+C ... (Ans.)
(ここで ln( ) は自然対数, , Cは任意定数。)
