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xArCtAn2xDx不定积分

∫x^2arctanxdx=1/3x^3arctanx-1/6x^2+1/6ln(1+x^2)+C.(C为积分常数)∫(x^2)*arctanxdx=1/3∫arctanxdx^3=1/3x^3arctanx-1/3∫x^3/(1+x^2)dx=1/3x^3arctanx-1/6∫x^2/(1+x^2)dx^2=1/3x^3arctanx-1/6∫[1-1/(1+x^2)]dx^2=1/3x^3arctanx-1/6x^2+1/6ln(1+x^2)+C

使用分部积分法,得到 ∫x arctan2x dx=1/2 *∫ arctan2x dx^2=x^2/2 *arctan2x - 1/2 *∫ x^2 *darctan2x=x^2/2 *arctan2x - ∫ x^2 /(1+4x^2)dx=x^2/2 *arctan2x - ∫1/4 +1/4 *1/(1+4x^2)dx=x^2/2 *arctan2x - x/4 -∫1/8 *1/(1+4x^2)d(2x)=x^2/2 *arctan2x - x/4 -1/8 *arctan2x +C,C为常数

∫arctan2xdx=x*arctan2x-∫x darctan2x =x arctan2x-∫2x/(1+4x^2)dx =x arctan2x-1/4∫1/(1+4x^2)d(1+4x^2) x arctan2x-1/4 ln(1+4x^2)+c

分部积分法: ∫arctan2xdx =xarctan(2x)-∫xd(arctan2x) =xarctan(2x)-∫2x/[1+(2x)^2]dx =xarctan(2x)+∫2x/[1+4x^2]dx =xarctan(2x)+2/4∫1/[1+4x^2]d(1+4x^2) =xarctan(2x)+1/2ln(1+4x^2)+C

∫xarctan2xdx = (1/2)∫arctan2xdx= (1/2)xarctan2x - (1/2)∫2xdx/(1+4x)= (1/2)xarctan2x - (1/4)∫(1+4x-1)dx/(1+4x)= (1/2)xarctan2x - x/4 +(1/8)∫d(2x)/(1+4x)= (1/2)xarctan2x - x/4 +(1/8)arctan2x + c (c为常数)

用分部积分法 我这有个关于 ∫xarctanxdx =∫arctanxd(0.5*x^2) =0.5*x^2 *arctanx-∫0.5*x^2d(arctanx) =0.5*x^2 *arctanx-∫0.5*x^2/(1+x^2)dx =0.5*x^2 *arctanx-0.5*∫(1-(1/(1+x^2))dx =0.5*x^2 *arctanx-0.5*∫dx+0.5*∫(1/(1+x^2))dx =0.5*x^2 *arctanx-0.5x+0.5*arctanx+C 你可以参照这个 再把2x代换就可以啦

∫x^2arctanxdx=(1/3)∫arctanxd(x^3)=(1/3)x^3.arctanx -(1/3)∫x^3/(1+x^2) dxconsiderx^3 = x(1+x^2) -x =x(1+x^2) -(1/2)(2x)∫x^2arctanxdx=(1/3)x^3.arctanx -(1/3)∫x^3/(1+x^2) dx=(1/3)x^3.arctanx -

设u = arctanx,(x^2)dx = v; 那么可以求出v = ∫[x^2/(1+x^2)]dx = x - arctanx; (分步积分)∫(x^2)arctanxdx= v*arctanx - (v'*arctanx)'= (x - arctanx)*arctanx - ∫[1-1/(x^2+1)]*[1/(x^2+1)]dx接下来就好做了

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