∵∫[x+(arctanx)²]/(1+x²)dx=∫[x/(1+x²)+(arctanx)²/(1+x²)]dx
=∫[x/(1+x²)]dx+∫[(arctanx)²/(1+x²)]dx
又∫[x/(1+x²)]dx=1/2∫d(1+x²)/(1+x²)
=1/2ln(1+x²)
∫[(arctanx)²/(1+x²)]dx=(arctanx)³-2∫[(arctanx)²/(1+x²)]dx (应用分部积分法)
上述关系式是一个关于∫[(arctanx)²/(1+x²)]dx的方程
解此方程得：∫[(arctanx)²/(1+x²)]dx=1/3(arctanx)³
∴∫[x+(arctanx)²]/(1+x²)dx=1/2ln(1+x²)+1/3(arctanx)³+C （C是积分常数）.