∫x/(x^4+2(x^2)+5)dx

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∫x/(x^4+2(x^2)+5)dx
∫x/(x^4+2(x^2)+5)dx

∫x/(x^4+2(x^2)+5)dx
令a=x²
原式=1/2*∫dx²/(x^4+2x²+5)
=1/2*∫da/(a²+2a+1+4)
=1/2*∫da/[(a+1)²+4]
=1/8*∫da/[(1/4)(a+1)²+1]
=1/8*∫da/[(a/2+1/2)²+1]
=1/4*∫d(a/2+1/2)/[(a/2+1/2)²+1]
=1/4*arctan((a/2+1/2)+C
=1/4*arctan(x²/2+1/2)+C

①（y+1)÷4=(x+2)÷3
两边乘以12
3(y+1)=4(x+2)
3y+3=4x+8
3y=4x+5
代入②2x-3y=1
2x-(4x+5)=1
2x-4x-5=1
2x=-6
x=-3
所以3y=4x+5=-7
所以x=-3,y=-7/3

∫x/(x^4+2(x^2)+5)dx ∫x/(x^2-4x-5)dx ∫ x/√(x^2+4x+5)dx ∫x/((x^2)+4x+5)dx ∫X√(2-5X)dx ∫x/（x^2+5）dx ∫（x^2+1/x^4)dx ∫dx/根号(4x-x^2) ∫1/(x^4-x^2)dx ∫x/(4+9x^2)dx ∫(x^3-x^2+x+1)/(x^2+1) dx∫(x+4)/(x^2-x-2) dx ∫4/((2-6X-X^2)^(5/2)) dx 计算不定积分∫dx/(4x^2+4x+5) ∫dx／4x^2+4x+5 求不定积分∫1/(x^2+4x+5)dx. ∫1/(x^2-5x+4)dx= 过程 ∫√（5-4x-x^2）dx ∫ dx/√(x^2+4x+5)