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
