求不定积分[x^2/(1+x^2)]arctanxdx

求不定积分[x^2/(1+x^2)]arctanxdx
求不定积分[x^2/(1+x^2)]arctanxdx

求不定积分[x^2/(1+x^2)]arctanxdx
∫[1-1/(1+x^2)]arctnxdx
=∫arctanxdx-∫arctanxdarctanx
=xarctanx-∫xdx/(1+x^2)-(1/2)(arctanx)^2+C
=xarctanx-(1/2)ln(1+x^2)-(1/2)(arctanx)^2+C

du =(u^2)arctanu-∫[1-1/(1+u^2+1] du =(u^2)arctanu-(u-arctanu)+C =xarctan√x-√x+arctan√x+C=(x+1)arcran√x-√x+C (4)