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xArCtAnxDx的不定积分

∫xarctanxdx=x/2arctanx-1/2x+1/2arctanx+c.c为积分常数.解答过程如下:∫xarctanxdx=∫arctanxdx/2=x/2arctanx-∫x/2darctanx=x/2arctanx-1/2∫x/(1+x)dx=x/2arctanx-1/2∫(x+1-1)/(1+x)dx=x/2arctanx-1/2∫1-1/(1+x)dx=x/2

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

∫xarctanxdx=1/2∫arctanxdx^2=x^2arctanx/2-1/2∫x^2darctanx=x^2arctanx/2-1/2∫x^2/(1+x^2)dx=x^2arctanx/2-1/2∫1-1/(1+x^2)dx=x^2arctanx/2-x/2+arctanx+C

∫ x * arctanx dx= ∫ arctanx d(x/2)= (x/2)arctanx - (1/2)∫ x d(arctanx)= (x/2)arctanx - (1/2)∫ x/(x + 1) dx= (x/2)arctanx - (1/2)∫ (x + 1 - 1)/(x + 1) dx= (x/2)arctanx - (1/2)∫ dx + (1/2)∫ dx/(x + 1)= (x/2)arctanx - x/2 + (1/2)arctanx + c

∫xarctanxdx=1/2∫arctanxd(x)=x/2arctanx-1/2∫x/(1+x)dx=x/2arctanx-1/2∫[1-1/(1+x)]dx=x/2arctanx-x/2+1/2arctanx+c=(x+1)/2arctanx-x/2++c

∫xarctanxdx=(1/2)∫ arctanxd(x)分部积分=(1/2)xarctanx - (1/2)∫ x/(1+x) dx=(1/2)xarctanx - (1/2)∫ (x+1-1)/(1+x) dx=(1/2)xarctanx - (1/2)∫ 1 dx + (1/2)∫ 1/(1+x) dx=(1/2)xarctanx - (1/2)x + (1/2)arctanx + C

∫xarctanxdx=(1/2)∫arctanxd(x^2)=(1/2)x^2arctanx-(1/2)∫x^2d(arctanx)=(1/2)x^2arctanx-(1/2)∫[x^2/(1+x^2)]dx=(1/2)x^2arctanx-(1/2)∫dx+(1/2)∫[1/(1+x^2)]dx=(1/2)x^2arctanx-(1/2)x+(1/2)arctanx+C

∫xarctanxdx 分部积分=(∫arctanxdx^2)/2=x^2arctanx|(0,1)/2 - ∫x^2darctanx/2=π/8 - ∫(x^2/1+x^2)dx/2=π/8 - ∫(1-1/(1+x^2))dx/2=π/8 - ∫dx/2 + ∫dx/(1+x^2)/2=π/8 - x/2|(0,1) + arctanx/2|(0,1)=π/8 - 1/2 + π/8=π/4 - 1/2 X_Q_T提醒的没错,而且我把分部积分的公式记错了,汗啊

∫xarctanxdx令u(x)=x2,v(x)=arctanx;∫xarctanxdx=0.5*∫v(x)d[u(x)]利用分部积分法可得

∫ ln(1+x)/√x dx= 2∫ ln(1+x)/(2√x) dx= 2∫ ln(1+x) d√x= 2ln(1+x) * √x - 2∫ √x dln(1+x),integration by part= 2(√x)ln(1+x) - 2∫ √x/(1+x) dx= 2(√x)ln(1+x) - 2∫ (2√x*√x)/[2√x*(1+x)] dx= 2(√x)ln(1+x) - 4∫ [1+(√x)-1]/[1+(√x)] d√x= 2(√x)ln(1+x) -

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