If ∫xtan−1⁡x1+x2dx=1+x2f(x)+Klog⁡x+x2+1+C then

∫x1+x2dx=12∫dtt=12⋅t1/2=t=1+x2                                                            [where 1 + x2 = t]we integrate by the parts, taking tan−1⁡x as the first function and x/1+x2 as the second.⇒∫xtan−1⁡x1+x2dx=tan−1⁡x1+x2−∫11+x2⋅1+x2dx=1+x2tan−1⁡x−∫dx1+x2=1+x2tan−1⁡x−log⁡x+x2+1+CThus f(x)=tan−1⁡x,K=−1.