作业帮不定积分一道题 Sdx/(x^4+1)
来源：学生作业帮 编辑：作业帮 分类：综合作业 时间：2024/05/20 02:41:44
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/[(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
一道高数题,求不定积分的：∫(1-x)/√(9-4x^2)dx 的不定积分.
一道不定积分题求解∫1/(√x+1)dx
求一道不定积分题∫[(根号1+lnx)/x]dx
一道不定积分题2∫dx/(e^x+1)^2 .》.《
求解一道高数不定积分题 1/√(1+x^4)dx .
还有一道题：求不定积分,被积函数是1/(1+x^4)
求一道不定积分题（x + 1/x）^1/2dx
一道求不定积分的题：根号下x/（1+三次根号下x）
一道求不定积分的题：x^3 / （x^2+1） dx = 应该是用换元法计算.
一道高数题：求不定积分∫(x^2+x^4)^1/2dx.
一道不定积分的题,
一道不定积分题