4

∫

0

dx

1+x2

dx
Let x = tan θ ⇒ θ = tan−1 x
dx = sec2 θ d θ
When x = 0 , θ = tan−1 (0) = 0
When x = 1, θ = tan−1 1 =

π

4

∴

4

∫

0

dx

1+x2

=

π 4

∫

0

sec2θ

1+tan2θ

dθ
=

π 4

∫

0

sec2θ

sec2θ

dθ
=

π 4

∫

0

dθ
= [θ]0

π

4

= [

π

4

−0]
=

π

4