不定积分x^7dx/(x^4+2) 不定积分(3x^4+x^3+4x^2+1)dx/(x^5+2x^3+x) 不定积分dx/(x^4+x^2+1）

Question

如题所述

Answer 1

∫x^7/(x⁴+2) dx
= ∫x³(x⁴+2-2)/(x⁴+2) dx
= ∫x³ dx - 2∫x³/(x⁴+2) dx
= x⁴/4 - 2(1/4)∫1/(x⁴+2) d(x⁴+2)
= x⁴/4 - (1/2)ln|x⁴+2| + C

∫(3x⁴+x³+4x²+1)/(x^5+2x³+x) dx
= ∫(3x⁴+x³+4x²+1)/[x(x²+1)²] dx
= ∫(2x+1)/(x²+1) dx - ∫1/(x²+1)² + ∫1/x dx
= 2∫x/(x²+1) dx + ∫1/(x²+1) dx - ∫1/(x²+1)² + ∫1/x dx
= ln(x²+1) + arctanx + ln|x| - ∫1/(x²+1)² dx
对于∫1/(x²+1)² dx，代入x=tanθ => dx=sec²θ dθ，sinθ=x/√(x²+1)，cosθ=1/√(x²+1)
= ∫cos²θ dθ = (1/2)∫(1+cos2θ) dθ = θ/2 + 1/2*sinθcosθ
= (1/2)arctanx + x/[2(x²+1)]
原式 = ln|(x²+1)x| + (1/2)arctanx - x/[2(x²+1)] + C

∫1/(x⁴+x²+1) dx，可设1/(x⁴+x²+1)=(Ax+B)/(x²+x+1)+(Cx+D)/(x²-x+1)，用待定系数法解出ABCD
= (1/2)∫(x+1)/(x²+x+1) dx - (1/2)∫(x-1)/(x²-x+1) dx
= [(1/4)∫(2x+1)/(x²+x+1) dx + (1/4)∫dx/(x²+x+1)] + [-(1/4)∫(2x-1)/(x²-x+1) dx + (1/4)∫dx/(x²-x+1)]
= (1/4)ln|x²+x+1| - (1/4)ln|x²-x+1| + (1/4)∫dx/[(x+1/2)²+3/4] + (1/4)∫dx/[(x-1/2)²+3/4]
= (1/4)ln| (x²+x+1)/(x²-x+1) | + 1/(2√3)*{arctan[(2x+1)/√3] + arctan[(2x-1)/√3]} + C
= (1/4)ln| (x²+x+1)/(x²-x+1) | + 1/(2√3)*arctan[√3x/(1-x²)] + C

Answer 2

∫x^7dx/(x^4+2)
=(1/4)∫x^4d(x^4)/(x^4+2)
=(1/4)x^4-(1/4)ln(x^4+2)+C

∫（3x^4+x^3+4x^2+1)dx/(x^5+2x^3+x)
=∫(3x^4+6x^2+3)dx/(x^5+2x^3+x)+∫(x^3-2x^2-2)dx/(x^5+2x^3+x)
=3∫dx/x+ ∫(x^3+x)dx/(x^5+2x^3+x)+∫-xdx/(x^5+2x^3+x) +∫(-2x^2-2)dx/(x^5+2x^3+x)
=3lnx+arctanx+∫-dx/(x^2+1)^2 +∫-2dx/x(x^2+1)
=3lnx+arctanx -∫dx/(x^2+1)^2+2∫(x^2-(x^2+1))dx/[x(x^2+1)]
=3lnx+arctanx-∫dx/(x^2+1)^2+2∫xdx/(x^2+1) -2∫dx/x
=lnx+arctanx-∫dx/(x^2+1)^2+ln(x^2+1)
∫dx/(x^2+1)^2 x=tanu ∫dx/(x^2+1)^2=∫secu^2du/secu^4=∫cosu^2du=u/2+sin2u/4
=arctanx /2+(1/2)x/ (1+x^2)
=lnx+arctanx/2 -(1/2)x/(1+x^2)+ln(x^2+1)+C

∫dx/(x^4+x^2+1)
=∫dx/[(x^2+x+1)(x^2-x+1)]
=(1/2)∫dx/[x(x^2-x+1)] -(1/2)∫dx/[x(x^2+x+1)]
∫dx/[x(x^2-x+1)] =∫[x^2-(x^2-x+1)]dx/[x(x-1)(x^2-x+1)
=∫xdx/[(x-1)(x^2-x+1)] - ∫dx/[x(x-1)]
=-∫(x-1)dx/(x^2-x+1)+∫dx/(x-1)-∫dx/(x-1)+∫dx/x
=(-1/2)ln(x^2-x+1)+(-1/2)∫dx/(x^2-x+1)+ln|x|
=(-1/2)ln(x^2-x+1)+(-1/(√3))arctan[(2x-1)/√3] +ln|x|+C

∫dx/[x(x^2+x+1)]=∫[(x^2+x+1)-x^2]dx/[x(x+1)(x^2+x+1)]
=∫dx/[x(x+1)]-∫xdx/[(x+1)(x^2+x+1)]
=ln|x|-ln|x+1| -∫[(x+1)^2-(x^2+x+1)]dx/[(x+1)(x^2+x+1)]
=ln|x|-ln|x+1|-∫(x+1)dx/(x^2+x+1)+∫dx/(x+1)
=ln|x|-(1/2)ln(x^2+x+1)-∫(1/2)dx/[(x+1/2)^2+3/4]
=ln|x|-(1/2)ln(x^2+x+1)-(1/√3)arctan[(2x+1)/√3]
代入
∫dx/(x^4+x^2+1)=(-1/4)ln(x^2-x+1)+(-1/(2√3))arctan[(2x-1)/√3]
+ (1/4)ln(x^2+x+1)+(1/(2√3))arctan[(2x+1)/√3]+C本回答被提问者采纳