1/(x^1/4+x^1/2)dx=
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1/(x^1/4+x^1/2)dx=
1/(x^1/4+x^1/2)dx=
1/(x^1/4+x^1/2)dx=
∫1/(x⁴+x²+1) dx
= (1/2)∫(x+1)/(x²+x+1) dx - (1/2)∫(x-1)/(x²-x+1) dx
= (1/2)[(1/2)∫(2x+1)/(x²+x+1) dx + (1/2)∫1/(x²+x+1) dx]
- (1/2)[(1/2)∫(2x-1)/(x²-x+1) dx - (1/2)∫1/(x²-x+1) dx]
= (1/4)∫d(x²+x+1)/(x²+x+1) - (1/4)∫d(x²-x+1)/(x²-x+1)
+ (1/4)∫d(x+1/2)/[(x+1/2)²+3/4] + (1/4)∫d(x-1/2)/[(x-1/2)²+3/4]
= (1/4)ln|(x²+x+1)/(x²-x+1)| + (2/√3)arctan[(2x+1)/√3] + (2/√3)arctan[(2x-1)√3] + C
1/(x+x^2)dx
∫(x^2+1/x^4)dx
∫1/(x^4-x^2)dx
∫ x/(1+X^2)dx=
∫(x+1/x)^2dx=?
∫cos x / ( 1 + (sinx)^2 ) dx = ∫x^3 / ( 1 + x^4 ) dx = ∫(sec x)^3 * tan x dx = ∫x^2 * e^(-2x) dx = ∫x * cos 2x dx = ∫(cos 2x)^2 dx = ∫(13x - 6) / ( x (x-2)(x+3) )dx = ∫(x^2 + 2x - 2) / ((x-2)(x+1)) dx = 请把过程写出来哈.>< 旷
这个怎恶魔算∫cos x / ( 1 + (sinx)^2 ) dx = ∫x^3 / ( 1 + x^4 ) dx = ∫(sec x)^3 * tan x dx = ∫x^2 * e^(-2x) dx = ∫x * cos 2x dx = ∫(cos 2x)^2 dx = ∫(13x - 6) / ( x (x-2)(x+3) )dx = ∫(x^2 + 2x - 2) / ((x-2)(x+1)) dx =
不定积分 [1/(e^x+e^(-x))]dx=?根号{[(x^2)-1]/x}dx=?
∫(1+x)/(X^2)dx=∫ [(1+x)/(X^2)]dx得什么?
1/(x^1/4+x^1/2)dx=
∫(1-x)/[√(9-4x^2)]dx=
∫1/(x^2-5x+4)dx= 过程
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∫(x^3-x^2+x+1)/(x^2+1) dx∫(x+4)/(x^2-x-2) dx
1.dx/dt = 4-2x;2.dx/dt = 1+x.
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∫(x+2)/(4x-x^2)^1/2dx