求这个定积分

解答过程如图：

∫(2x+1)/(x^2+1)dx 上限1 下限-1
=∫[2x/(x^2+1)]dx+∫[1/(x^2+1)]dx (上限1 下限-1)
=0+2∫[1/(x^2+1)]dx (上限1 下限0)
(这里用到了奇偶函数在对称区间上的定积分的结论)
=2arctanx (在1处的值减 在0处的值)
=π/2

只给提示
∫(2x+1)/(x^2+1)dx
=∫2x/(x^2+1)dx+∫1/(x^2+1)dx
=∫1/(x^2+1)d(x^2+1)+∫1/(x^2+1)dx
=...

∫(2x+1)dx/(x^2+1)=∫2xdx/(x^2+1)+∫dx/(x^2+1)=∫d(x^2+1)/(x^2+1)+arctan(x)=ln(x^2+1)+arctan(x)=ln(3)+arctan(1)-ln(3)-arctan(-1)=2arctan(1)=