Question

Derive the integral for $\frac{1}{1+x^2}$

Answer 1

Let say

$I=\int \frac{1}{1+x^2}dx------\left(1\right)$

Now, Put

$$\begin{gathered} x=\tan t \\ \Rightarrow t={\tan }^{-1}x \\ Differentiatingw.r.txweget \\ \Rightarrow 1=se{c}^{2}t\frac{dt}{dx} \\ \Rightarrow dx=se{c}^{2}tdt \end{gathered}$$

Putting in equation (1) we get

$\begin{array}{rcl}I& =& \int \frac{1}{1+x^2}dx=\int \frac{se{c}^{2}t}{1+{\tan }^{2}t}dt\\ & \Rightarrow & I=\int \frac{se{c}^{2}t}{se{c}^{2}t}dt=\int 1dt\left[\because 1+{\tan }^{2}t=se{c}^{2}t\right]\\ & \Rightarrow & I=t+C\\ & \Rightarrow & I={\tan }^{-1}x+C\end{array}$