Question

#HLWRC高数#求解不定积分∫xe^arctanx/((x^2+1)^(3/2))dx，#高等数学#最佳分部积分法需要移项？

#HLWRC高数#求解不定积分∫xe^arctanx/((x^2+1)^(3/2))dx，#高等数学#最佳分部积分法需要移项。

Answer 1

Answer 2

令 x = tanu
I = ∫xe^arctanx/[(x^2+1)^(3/2)]dx = ∫tanu(secu)^2e^udu/(secu)^3
= ∫tanue^udu/secu = ∫sinue^udu = ∫sinude^u
= e^usinu - ∫e^ucosudu = e^usinu - ∫cosude^u
= e^u(sinu-cosu) - I
I = (1/2)e^u(sinu-cosu) + C = (1/2)[(x-1)/√(1+x^1)]e^arctanx + C