∫x 根号(1-2x^2)dx
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=-1/2∫√(1-x^2)d(1-x^2)=-1/2×2/3√(1-x^2)^3+C=-1/3√(1-x^2)^3+C
/>令x=sect,则dx=sect·tantdt∫(1→2)√(x²-1)/xdx=∫(0→π/3)tan²tdt=∫(0→π/3)(sec²t-1)dt=(tant-
∫sin²xcos²xdx=∫(1/2·2sinxcosx)²dx=(1/4)∫sin²2xdx=(1/4)∫(1-cos4x)/2dx=(1/8)(x-1/4
设x=tanα则√(x²+1)=1/cosα∴原式=∫d(tanα)/(tanα+1/cosα)=∫(1/cos²α)/(tanα+1/cosα)dα=∫(cosα)dα/(sin
∫1/根号x*sec^2(1-根号x)dx=2∫sec^2(1-根号x)d(√x)=-2∫sec^2(1-根号x)d(1-√x)=-2tan(1-√x)+c
∫dx/√(4x-x^2)=∫dx/√([4-(x-2)^2]=arcsin[(x-2)/2]+C
令t=√x∫1/(1+2√x)dx=∫1/(1+2t)dt^2=∫2t/(1+2t)dt=∫1-1/(1+2t)dt=∫dt-∫1/(1+2t)dt=t+1/2ln(1+2t)+C=√x+1/2ln(
∫[3^(1/x)/x²]dx=-∫3^tdt……t=1/x=-(3^t)/ln3+C=-[3^(1/x)]/ln3+C;∫√x/(√x-1)dx=∫[1+1/(√x-1)]dx=x+∫2√
∫dx/x[根号1-(ln^2)x]=∫d(lnx)/[根号1-(ln^2)x]=∫dt/[根号1-t^2](设t=lnx)=arcsint+C=arcsin(lnx)+C
答:∫xf(x)dx=x/√(1-x²)+C两边求导得:xf(x)=1/√(1-x²)+(-x/2)*(-2x)/[(1-x²)√(1-x²)]=(1-x
设√(5-4x)=yx=(5-y²)/4dx=-ydy/2则∫x/√(5-4x)dx=∫(5-y²)(-ydy/2)/4y=∫(y²-5)dy/8=y³/24-
原式=∫1/(1-x)(1+x)dx=1/2∫[1/(1-x)+1/(1+x)]dx=1/2[-ln|1-x|+ln|1+x|]+c=1/2ln|(1+x)/(1-x)|+c啊,原来有根号啊应该是ar
(x^2)/2-18x^(1/2)+3x+C0.5*x^2+2*x^(1/2)+C9x-2x^3+0.2*x^5+C
再答:满意的话请采纳一下再答:满意的话请采纳一下再问:根号1+tant^2应该是1/cost再答:我错了再答:再答:再问:3Q再问:dx/2x^2+3x-2再问:曲线y=1/2x^2上有一点M,该点处
解∫x√(4x²-1)dx=1/8∫√(4x²-1)d(4x²-1)=1/8∫√udu=1/8×(2/3)×u^(3/2)+C=1/12(4x²-1)^(3/2
原式=∫dx/(2X-1)^3/2=1/2∫(2X-1)^(-3/2)d(2x-1)=-根号(2x-1)
∫(2-x)/√(1-x^2)dx=2∫1/√(1-x^2)dx-∫x/√(1-x^2)dx=2arcsinx+√(1-x^2)+c