2x-1 x 2的不定积分
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∫[2x/(x^2+x+1)]dx=∫[(2x+1)/(x^2+x+1)]dx-∫dx/(x^2+x+1)=ln|x^2+x+1|-∫dx/(x^2+x+1)considerx^2+x+1=(x+1/
∫1/(x²-x-2)dx=∫1/[(x-2)(x+1)]dx=1/3∫[1/(x-2)-1/(x+1)]dx=1/3[∫1/(x-2)dx-∫1/(x+1)dx]=1/3[ln(x-2)-
∫(lnx-1)/x²dx=-∫(lnx-1)d(1/x)=-[(lnx-1)/x-∫1/xd(lnx-1)]=-(lnx-1)/x+∫1/x²dx=-(lnx-1)/x-1/x+
答:1.∫arcsinxdx可用分部积分原式=xarcsinx-∫x/√(1-x^2)dx=xarcsinx+√(1-x^2)+C2.∫e^(√x+1)dx换元,令√(x+1)=t,则x=t^2-1,
∫arctanxdx/[x^2(1+x^2)]=∫arctanxdx/x^2-∫arctanxdx/(1+x^2)=∫arctanxd(-1/x)-∫arctanxdarctanx=-(arctanx
1/(x+1)(x+2)(x+3)=1/(x+1)[1/(x+2)-1/(x+3)]=1/[(x+1)(x+2)]-1/[(x+1)(x+3)]=1/(x+1)-1/(x+2)-1/2[1/(x+1)
∫xlnx/(1+x^2)^2dx=1/2*∫lnx/(1+x^2)^2d(1+x^2)=-1/2*∫lnxd[1/(1+x^2)]=-1/2*lnx*1/(1+x^2)+1/2*∫[1/(1+x^2
令1/x=t则原式=∫arctant/(1+1/t²)*(-1/t²)dt=∫-arctant/(1+t²)dt=∫-arctantdarctant=-1/2arctan
配方:1+x-x^2=5/4-(x-1/2)^2,套用不定积分公式(∫dx/√(a^2-x^2))结果是arcsin((2x-1)/√5)+C
令x=tant,t∈(-π/2,π/2),则√(1+x²)=sect,dx=sec²tdt∫√(1+x²)dx=∫sec³tdt=∫sectd(tant)=se
令x=siny原式=∫1/(sinycosy)*cosydy=∫1/[2cos^2(y/2)]/tan(y/2)dy=∫d(tany/2)/tan(y/2)=ln|tan(y/2)|+C=ln|(1-
∫dx/(x^2+x)=∫[1/x-1/(x+1)]dx=ln|x/(x+1)|+C
1/(1+x^2)d(1+x^2)=ln(1+x^2)+C
∫x/(x^2+x+1)dx=(1/2)∫dln(x^2+x+1)-(1/2)∫1/(x^2+x+1)dx=(1/2)ln(x^2+x+1)-(1/2)∫1/(x^2+x+1)dxx^2+x+1=(x
∫[(x-1)/(x^2+3)]dx=∫[x/(x^2+3)]dx-∫[1/(x^2+3)]dx=(1/2)∫[1/(x^2+3)]d(x^2+3)-(1/√3)∫{1/[(x/√3)^2+1]}d(
再答:诚邀您加入百度知道团队“驾驭世界的数学”。
分部积分法再答:
(-(x/(1+x^2))+ArcTan[x])/2