Derive the integral for 1/1 + x2

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

Solution

Let us assume that

I=

\(\begin{array}{l}\int \frac{1}{1+x^{2}}\end{array} \)

Substitute x= tan u

dx=sec2udu

u = arctan x———-(i)

So

I=

\(\begin{array}{l}\int 1/ 1 + tan^{2}u . sec^{2}u du\end{array} \)

=

\(\begin{array}{l}\int 1/sec^{2}u . sec^{2}u du\end{array} \)

=

\(\begin{array}{l}\int 1.du\end{array} \)

=u

= arctan x (from (i))

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

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