Question

反常积分计算求解

如图

Answer 1

设x=t^4,dx=4t³dt
原积分=
4∫（0，+∞）t^4/(1+t^4)².dt
=4[∫（0，+∞）dt/(1+t^4)-∫（0，+∞）dt/(1+t^4)²]
∫dt/(1+t^4)=√2/8{ln[(x²+√2x+1）/[(x²-√2x+1）]+2[arctan(√2x+1）+arctan(√2x-1）]}
∫dt/(1+t^4)²=1/32×{3√2ln[(x²+√2x+1）/[(x²-√2x+1）]+6√2[arctan(√2x+1）+arctan(√2x-1）]+8x/(x^4+1)}
ln[(x²+√2x+1）/[(x²-√2x+1）]|(0，+∞）=0
arctan(√2x+1）+arctan(√2x-1）]|(0，+∞）=π/2-π/4+π/2+π/4=π
x/(x^4+1)|(0，+∞）=0
代入化简得：
4×√2/8×2π-4×1/32×[6√2 π]
=(√2-6√2/8）π
=√2π/4