令t＝x＋1,则x＝t－1,dx＝dt
∫(3x＋5)dx/(x^2＋2x＋2)^2
＝∫(3t＋2)dt/(t^2＋1)^2 分部积分法
＝－∫(3t＋2)/2t d[1/(t^2＋1)]
＝－3/2∫d[1/(t^2＋1)]－∫1/t d[1/(t^2＋1)]
＝－3/2(t^2＋1)－1/t×1/(t^2＋1)－∫1/t^2×1/(t^2＋1)dt
＝－3/2(t^2＋1)－1/t×1/(t^2＋1)－∫[1/t^2－1/(t^2＋1)]dt
＝－3/2(t^2＋1)－1/t×1/(t^2＋1)＋1/t＋arctant＋C
＝(2t－3)/[2(t^2＋1)]＋arctant＋C
＝(2t－1)/2(x^2＋x＋2)＋arctan(x＋1)＋C