不定积分一道题 Sdx/(x^4+1)
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/20 00:15:00
不定积分一道题 Sdx/(x^4+1)
不定积分一道题 Sdx/(x^4+1)
不定积分一道题 Sdx/(x^4+1)
∫dx/(x^4+1)=∫dx/[(x^2+1)^2-2x^2]
=∫dx/[(x^2+1-√2x)(x^2+1+√2x)]
=∫(1/2√2x)[ (x^2+1+√2x)-(x^2+1-√2x)]dx/[(x^2+1+√2x)(x^2+1-√2x)]
=∫(1/(2√2x))dx/(x^2+1-√2x) - ∫(1/(2√2x))dx/(x^2+1+√2x)
=(1/(2√2))[∫(1/2)d(x^2+1-√2x)/(x^2+1-√2x) +∫(1/2)*√2dx/(x^2+1-√2x)
-∫(1/2)d(x^2+1+√2x)/(x^2+1+√2x)+∫(1/2)*√2dx/(x^2+1+√2x)]
=(1/(2√2))*[(1/2)(ln|(x^2+1-√2x)|-ln|(x^2+1+√2x)|)
+(1/4)∫dx/[(x-√2/2)^2+1/2] +(1/4)∫dx/[(x+√2/2)^2+1/2]]
=(√2/4)[(1/2)(ln|(x^2+1-√2x)| -ln|(x^2+1+√2x)|)
+(√2/4)arctan(√2x-1)
+(√2/4)arctan(√2x+1)]+C
=(√2/8)[ln|(x^2+1-√2x)| -ln|(x^2+1+√2x)|)]
+(1/8)arctan(√2x-1)
+(1/8)arctan(√2x+1)+C
∫ dx/(x⁴+1)
= (1/2)∫ [(x²+1)-(x²-1)]/(x⁴+1) dx
= (1/2)∫ (x²+1)/(x⁴+1) dx - (1/2)∫ (x²-1)/(x⁴+1) dx
= (1/2)∫ (1+1/x²)/(x²+1/x²) dx...
全部展开
∫ dx/(x⁴+1)
= (1/2)∫ [(x²+1)-(x²-1)]/(x⁴+1) dx
= (1/2)∫ (x²+1)/(x⁴+1) dx - (1/2)∫ (x²-1)/(x⁴+1) dx
= (1/2)∫ (1+1/x²)/(x²+1/x²) dx - (1/2)∫ (1-1/x²)/(x²+1/x²) dx
= (1/2)∫ d(x-1/x)/[(x-1/x)²+2] - (1/2)∫ d(x+1/x)/[(x+1/x)²-2]
= (1/2) * (1/√2) * arctan[(x-1/x)/√2] - (1/2) * 1/(2√2) * ln|(x+1/x-√2)/(x+1/x+√2)| + C
= 1/(2√2) * arctan[x/√2-1/(x√2)] - 1/(4√2) * ln|(x²-x√2+1)/(x²+x√2+1)| + C
收起