Let `u=tan^(-1)((sqrt(1+x^(2))-1)/(x)) and v=tan^(-1)x`
Now, `v=tan^(-1)x rArr x = tanv.`
Putting `x=tan v,` we get
`u=tan^(-1){(sqrt(1+tan^(2)v)-1)/(tanv)}=tan^(-1)((secv-1)/(tanv))`
`=tan^(-1)((1-cosv)/(sinv))=tan^(-1){(2sin^(2)(v//2))/(2sin(v//2)cos(v//2))}`
`=tan^(-1){tan.(v)/(2)}=(v)/(2).`
`thereforeu=(v)/(2)rArr(du)/(dv)=(1)/(2).`