The correct option is B π4
I=1∫−1dx(1+ex)(1+x2)⋯(i)
Using the property: b∫af(x) dx=b∫af(a+b−x) dx
=1∫−1dx1+e−x.11+x2
I=1∫−1exdx(1+ex)(1+x2)⋯(ii)
Adding (i) and (ii), we get
2I=1∫−1(1+ex)dx(1+ex)(1+x2)
=1∫−1dx(1+x2)
We know that,
a∫−af(x) dx=⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩2a∫0f(x) dx, if f(x) is even0, if f(x) is odd
=21∫0dx(1+x2)
I=1∫0dx(1+x2)=tan−1(1)=π4