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∫1/x∧2Dx=

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

如下图

因为这里书写不便,故将我的答案做成图像贴于下方,谨供楼主参考(若图像显示过小,点击图片可放大)。

令u=1+x^2 则du=2xdx 原式=1/2·∫1/udu =1/2·lnu+C =1/2·ln(1+x^2)+C

令x=tan(t), 则dx=(sect)^2dt, 带入∫(1+x^2)^(1/2)dx =∫sectdtant =secttant-∫tantdsect =sect*tant-∫sect*tan²tdt =sect*tant-∫sect(sec²t-1)dt =secttant-∫sec³tdt+∫sectdt =secttant-∫sec³tdt+ln|sect+tant| 2∫sec³...

郭敦荣回答: 若换元时,则应换到底;若不换元,则积分上下限不变。 换元法,令u=1+x²,则du=d(1+x²)=2xdx。x=0时,u=1;x=1时,u=2, 于是,原式=(1/2)∫(1,2)(1/u)du=(1/2) lnu|(1,2) =(1/2)(ln2-ln1) =(1/2)ln...

收敛的

∫(0→2) x√(2x - x2) dx = ∫(0→2) x√[1 - (x - 1)2] dx 令x - 1 = sinθ,dx = cosθdθ,第二换元积分法当x = 0,θ = - π/2 当x = 2,θ = π/2 = ∫(- π/2→π/2) (1 + sinθ) * cos2θ dθ = ∫(- π/2→π/2) cos2θ dθ + ∫(- π/2→π/2) sinθcos2θ dθ,第一个偶函数...

∫ (1-x-x²)/(x²+1)² dx =∫ (2-x-x²-1)/(x²+1)² dx =2∫ 1/(x²+1)² dx - ∫ x/(x²+1)² dx - ∫ (x²+1)/(x²+1)² dx =2∫ 1/(x²+1)² dx - (1/2)∫ 1/(x²+1)² dx&...

令x=asint dx=acostdt t=arcsin(x/a) 原式=∫a^2cos^2tdt =a^2/2*∫(1+cos2t)dt =a^2/2*(t+sintcost)+C =a^2/2*arcsin(x/a)+ax/2*(1-x^2/a^2)^(1/2)+C

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