# Derive the integral for 1/1 + x2

We need to evaluate the integral of 1/(1+x2)

### Solution

Let us assume that

I=$$\int \frac{1}{1+x^{2}}$$

Substitute x= tan u

dx=sec2udu

u = arctan x———-(i)

So

I= $$\int 1/ 1 + tan^{2}u . sec^{2}u du$$

= $$\int 1/sec^{2}u . sec^{2}u du$$

=$$\int 1.du$$

=u

= arctan x (from (i))

So the integral of 1/(1+x2) is arc tanx