The correct option is
A π4−12log2taking u=tan−1x,v=1
integrating by parts we get ∫1×tan−1xdx=tan−1x∫1dx−∫(11+x2∫1dx)dx=xtan−1x−∫x1+x2dx=xtan−1x−12∫2x1+x2dx=xtan−1x−12ln1+x2
∫10tan−1xdx=[xtan−1x−12ln1+x2]10=[π4−12ln2]−0=π4−12ln2
so A is the correct option