Question

高数∫xsin(x^4)dx 求这个不定积分过程，这个无初等函数表示的这个

Answer 1

解答：
∫[1/（1+x^4）]dx
= 1/2∫[(x^2+1)-(x^2-1)]/（1+x^4）dx
= 1/2 {∫(x^2+1)/(1+x^4) dx - ∫(x^2-1)/（1+x^4）dx }
= 1/2 {∫(1+1/x^2)dx /（x^2+1/x^2） - ∫(1-1/x^2)dx/（x^2+1/x^2）}
= 1/2 {∫d(x-1/x) /[(x-1/x)^2+2] - ∫d(x+1/x) /[(x+1/x)^2 -2] }
= 1/2 { 1/√2 ∫d[(x-1/x) /√2] /{[(x-1/x)/√2]^2+1} - ∫d(x+1/x) /[(x+1/x)^2 -2] }
= 1/2 { 1/√2 ∫d[(x-1/x) /√2] /{[(x-1/x)/√2]^2+1}
- 1/2√2 ∫d[(x+1/x) /√2] [ 1/{[(x+1/x)/√2] -1} - 1/{[(x+1/x)/√2] +1 }]
= √2/4*arctan[(x-1/x)/√2] - √2/8*ln|(x^2-x√2+1)/(x^2+x√2 +1)| + C

追答

滚！！

追问

这种答案也好意思拿出来

Answer 2

设a=x^2
da=2xdx
∫xsin(x^4)dx =1/2∫sin(a^2)da
=asin(a^2)-∫sin(a^2)da
asin(a^2)=3/2∫sin(a^2)da
∫sin(a^2)da=2/3asin(a^2)
∫xsin(x^4)dx =1/3asin(a^2)
=1/3(x^2)sin(x^4)

追问

……我都把答案放出来了，你太天真了

Answer 3

∫[1/（1+x^4）]dx

= 1/2∫[(x^2+1)-(x^2-1)]/（1+x^4）dx

= 1/2 {∫(x^2+1)/(1+x^4) dx - ∫(x^2-1)/（1+x^4）dx }

= 1/2 {∫(1+1/x^2)dx /（x^2+1/x^2） - ∫(1-1/x^2)dx/（x^2+1/x^2）}

= 1/2 {∫d(x-1/x) /[(x-1/x)^2+2] - ∫d(x+1/x) /[(x+1/x)^2 -2] }

= 1/2 { 1/√2 ∫d[(x-1/x) /√2] /{[(x-1/x)/√2]^2+1} - ∫d(x+1/x) /[(x+1/x)^2 -2] }

= 1/2 { 1/√2 ∫d[(x-1/x) /√2] /{[(x-1/x)/√2]^2+1}
- 1/2√2 ∫d[(x+1/x) /√2] [ 1/{[(x+1/x)/√2] -1} - 1/{[(x+1/x)/√2] +1 }]

= √2/4*arctan[(x-1/x)/√2] - √2/8*ln|(x^2-x√2+1)/(x^2+x√2 +1)| + C

追问

乱复制