∫−aa logx+x2+1dx=
2 log a2+1
2 log a2+1−a
0
2 log a+a2+1
I=∫−aa logx+x2+1dx=∫−aa log−x+x2+1dx∫−aa f(x)dx=∫−aa f(−x)dx∫−aa log−x2+x2+1x+x2+1dx=−∫−aa logx+x2+1dx=−I⇒I=0