How do you integrate 1/(1+x^2)?

Question

____1____
( 1 + x^2 )

Answer 1

∫1/(1+x^2) dx
= arctan(x)
-----
Reason:
d arctan(x)/dx
= 1/(1+x^2)

Proof:
y = arctan(x)
tan(y) = x

Differentiate with respect to x,
sec^2(y)y' = 1

y'
= 1/sec^2(y)
= 1/[1+tan^2(y)]
= 1/(1+x^2)

Answer 2

A recognised standard integral is :-

I = ∫1/(a² + x²) dx = (1/a) tan^(-1)(x/a) + C

In this case a = 1 so I = tan^(-1) x + C

Answer 3

1st step :

let x = tan y --> y=arc tan x
dx = sec^2 y dy

int 1/(1+ x^2) dx = int 1/(1+tan^2 y) sec^2y dy
= int sec^2 y/sec^2 y dy
= int 1 dy
= y+c
= arc tan x +c