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 arctan√x - √x + arctan√x +C
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 arctan√x - √x + arctan√x +C