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用分部积分法求∫(1,0)ArCtAnxDx

原式= x arctan√x - ∫x d (arctan√x) 令t=√x,则 ∫x d (arctan√x) = ∫ t^2 d (arctant) = ∫ t^2 / (1+ t^2) dt = ∫ (t^2+1-1) / (1+ t^2) dt = ∫ 1 dt - ∫ 1 / (1+ t^2) dt = t - arctan t + C 将t=√x带入 = √x - arctan√x +C 所以原式= x arcta...

2)3)4)答案同楼上, 1)∫xsin2xdx=(-1/2)∫xdcos2x=(-1/2)xcos2x+(1/2)∫cos2xdx=(-1/2)xcos2x+(1/4)sin2x+C 2)∫xlnxdx=(1/2)∫lnxdx^2=(1/2)x^2lnx-(1/2)∫xdx=(1/2)x^2lnx-(1/4)x^2+C 3)∫arccosxdx=xarccosx-∫-xdx/√(1-x^2) =xarctanx-(1/2)d...

原式=∫arctanxdx-∫arctanxdx/(1+x²) ∫arctanxdx=xarctanx-∫xdx/(1+x²) =xarctanx-1/2*∫d(1+x²)/(1+x²) =xarctanx-ln√(1+x²)+C ∫arctanxdx/(1+x²)=∫arctanxd(arctanx) =arctan²x-∫arctanxdx/(1+x²) ∴∫ar...

∫xcos²xdx=∫x(1+cos2x)/2dx=1/2(∫xdx+∫xcos2xdx) =1/2(1/2x²+∫xcos2xdx) =1/2(1/2x²+1/2∫xdsin2x) =1/2(1/2x²+1/2(xsin2x-∫sin2xdx)) =1/2(1/2x²+1/2xsin2x+1/4cos2x)+C

把x先凑成x^2/2运用分布积分,之后的形式你就能看出来怎么做了

您好,答案如图所示:

可以吧,;令z = arctanx,tanz = x,dx = (secz)^2 dz ∫ arctanx dx = ∫ z * (secz)^2 dz = ∫ z d(tanz) = ztanz - ∫ tanz dz = ztanz + ln|cosx| + C = arctanx * x + ln|[1/√(1 + x^2)] + C = xarctanx - (1/2)ln(1 + x^2) + C 直接做也可以...

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