一道不定积分题求∫(x-1)arctanxdx

∫xarctanxdx
=∫arctanxd(0.5*x^2)
=0.5*x^2 *arctanx-∫0.5*x^2d(arctanx)
=0.5*x^2 *arctanx-∫0.5*x^2/(1+x^2)dx
=0.5*x^2 *arctanx-0.5*∫(1-(1/(1+x^2))dx
=0.5*x^2 *arctanx-0.5*∫dx+0.5*∫(1/(1+x^2))dx
=0.5*x^2 *arctanx-0.5x+0.5*arctanx+C
=0.5*x^2 *arctanx-0.5x+0.5*arctanx+C

∫ arctanx dx
= x * arctanx - ∫ x d(arctanx)
= x * arctanx - ∫ x/(1+x²) dx
= x * arctanx - (1/2)∫ d(x²)/(1+x²)
= x * arctanx - (1/2)∫ d(1+x²)/(1+x²)
= x * arctanx - (1/2)ln(1+x²) + C

合在一起你会了吧