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2xArCtAnx平方的不定积分

∫ 2x*arctanx dx= ∫ arctanx d(x^2)= x^2*arctanx - ∫ x^2 d(arctanx)= x^2*arctanx - ∫ x^2/(1 + x^2) dx= x^2*arctanx - ∫ (x^2 + 1 - 1)/(1 + x^2) dx= x^2*arctanx - ∫ [1 - 1/(1 + x^2)] dx= x^2*arctanx 旦尝测妒爻德诧泉超沪- x + arctanx + c= - x + (x^2 + 1)arctanx + c

分部积分法:原式=-1/xarctanx-∫(-1/x)1/(1+x^2)dx=-1/xarctanx+∫1/x(1+x^2)dx=-1/xarctanx+∫dx[1/x-x/(1+x^2)]dx=-1/xarctan+ln|x|-∫xdx/(1+x^2)=-1/xarctanx+ln|x|-0.5∫d(x^2)/(1+x^2)=-1/xarctanx+ln|x|-0.5ln(1+x^2)+c

∫( xarctanx)dx= (1/2)∫arctan(x) d(x^2)=x^2arctan(x)/2 - (1/2)∫ x^2dx/(1+x^2)=x^2arctan(x)/2 - (1/2)∫dx + (1/2)∫dx/(1+x^2)=x^2arctan(x)/2 - (x/2) + arctan(x)/2 + C,其中,C为任意常数.

y=2xarctan(y/x) (y/x)'=(y'/x-y/x^2) (arctanu)'=1/(1+u^2)y'=2arctan(y/x)+2x*(y'/x -y/x^2)*[1/(1+(y/x)^2)] 1) y'[1-2x^2/(x^2+y^2)]=2arctan(y/x)-2(y/x)*[x^2/(x^2+y^2)] y'=[2arctan(y/x)-2xy/(x^2+y^2)]/[1-2x^2/(x^2+y^2)]

∫ x(arctanx) dx= ∫ (arctanx) d(x/2)= (1/2)x(arctanx) - (1/2)∫ x * 2(arctanx) * 1/(1 + x) dx,分部积分法= (1/2)x(arctanx) - ∫ x/(1 + x) * arctanx dx=

分部积分法 ∫arctanxdx =x*arctanx-∫xd(arctanx) =xarctanx-∫x/(1+x^2)dx =xarctanx-1/2∫1/(1+x^2)d(x^2) =xarctanx-1/2ln(1+x^2)+c

∫ x arctan(x)/(1 + x) dx= (1/2)∫ arctan(x)/[1 + (x)] d(x)= (1/2)∫ arctan(x) d[arctan(x)]= (1/2) [arctan(x)]/2 + C= (1/4)[arctan(x)] + C 都只是些凑微分的简单步骤:x dx = d(x/2)1/(1 + x) dx = d(arctanx)

令 x^(1/2) =t , x = t^2 ,0≤t≤1 ∫(0,1) xarctanx^(1/2)dx= ∫(0,1) t^2 arctant d(t^2)= 1/2 ∫(0,1) arctant d(t^4)=[1/2 t^4 arctant ] (0,1) - 1/2 ∫(0,1) t^4 d(arctant)=π/8 - 1/2 ∫(0,1) t^4/(1+t^2) dt=π/8 - 1/2 ∫(0,1) (t^4-1+1)/(1+t^2) dt=π/8 - 1/2 ∫(0,1) t^2 - 1 + 1/(1+t^2)

∫xf''(x)dx=∫xdf'(x)=xf'(x)-∫f'(x)dx=xf'(x)-f(x)+C证明题设f(x)=2xarctanx-ln(1+x2)f'(x)=2arctanx-2x/(1+x2)+2x/(1+x2)=2arctanxf'(x)>0时,2arctanx>0得出x>0∴f(x)在[0,+∞)单调递增∴当x≥0时,f(x)≥f(0)=0,

∫(1+x^2-1)arctanx/(1+x^2)dx=∫(1+x^2)arctanx/(1+x^2)dx-∫arctanx/(1+x^2)dx=∫arctanxdx-∫arctanx/(1+x^2)dx=xarctanx-∫x/(1+x^2)dx-∫arctanxd(arctanx)=xarctanx-1/2∫1/(1+x^2)d(1+x^2)-1/2(arctanx)^2=xarctanx-1/2ln(1+x^2)-1/2(arctanx)^2+C

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