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

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