(1+x2)

dy

dx

+ y = tan−1 x

dy

dx

+

y

1+x2

=

tan−1x

1+x2

... (i)
Given equation is linear with
So, I.F. = e∫

1

1+x2

dx = etan−1x
Solution of (i)
yetan−1x = ∫ etan−1x(

tan−1x

1+x2

) dx ... (ii)
For R.H.S,let tan−1 x = t ⇒

1

1+tx2

dx = dt
By substituting in equation(ii)
yetan−1x = ∫ et . tdt
⇒ yetan−1x = [tet−et] + C
⇒ yetan−1x = etan−1x (tan−1x−1) + C
⇒ y = tan−1x−1+Ce−tan−1x