求下列不定积分,∫xarctanxdx
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求下列不定积分,∫xarctanxdx
∫xarctanxdx
=(1/2)∫arctanxd(x^2)
=(1/2)x^2·arctanx-(1/2)∫x^2d(arctanx)
=(1/2)x^2·arctanx-(1/2)∫[x^2/(1+x^2)]dx
=(1/2)x^2·arctanx-(1/2)∫dx+(1/2)∫[1/(1+x^2)]dx
=(1/2)x^2·arctanx-(1/2)x+(1/2)arctanx+C
再问: 求下列不定积分,∫x√x -1dx
再答: ∫[x√(x-1)]dx
=∫[(x-1+1)√(x-1)]dx
=∫(x-1)√(x-1)d(x-1)+∫√(x-1)d(x-1)
=(2/5)(x-1)^(5/2)+(2/3)(x-1)^(3/2)+C
=(2/5)(x-1)^2·√(x-1)+(2/3)(x-1)√(x-1)+C。
=(1/2)∫arctanxd(x^2)
=(1/2)x^2·arctanx-(1/2)∫x^2d(arctanx)
=(1/2)x^2·arctanx-(1/2)∫[x^2/(1+x^2)]dx
=(1/2)x^2·arctanx-(1/2)∫dx+(1/2)∫[1/(1+x^2)]dx
=(1/2)x^2·arctanx-(1/2)x+(1/2)arctanx+C
再问: 求下列不定积分,∫x√x -1dx
再答: ∫[x√(x-1)]dx
=∫[(x-1+1)√(x-1)]dx
=∫(x-1)√(x-1)d(x-1)+∫√(x-1)d(x-1)
=(2/5)(x-1)^(5/2)+(2/3)(x-1)^(3/2)+C
=(2/5)(x-1)^2·√(x-1)+(2/3)(x-1)√(x-1)+C。
求ln(1+x)/√xdx的不定积分 还有一题xarctanxdx的不定积分
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