∫arctanx/{x[√（1－x²）]} dx
=-√(1-x²)arctanx+∫√(1-x²)/(1+x²)dx.
設(shè)x=cost,則dx=-sintdt.
∴∫√(1-x²)/(1+x²)dx
=-∫sin²t/(1+cos²t)dt
=∫(cos²t-1)/(1+cos²t)dt
=∫(1-2/(1+cos²t))dt
=t-2∫sec²t/(2+tan²t)dt
=t-2∫1/(2+tan²t)d(tant)
=t-√2∫1/(1+tan²t/2)d(tant/√2)
=t-√2arctan(tant/√2)+C
=arccosx-√2arctan(√(1-x²)/(√2x))+C,(C是積分常數(shù)).
故∫arctanx/{x[√（1－x²）]} dx
=-√(1-x²)arctanx+arccosx-√2arctan(√(1-x²)/(√2x))+C,