# What is the integral of 1/(1+x^2)?

Jun 28, 2018

$\int \frac{1}{1 + {x}^{2}} \mathrm{dx} = {\tan}^{-} 1 x + C$

#### Explanation:

color(blue)(int(du)/(1+u^2)=tan^-1u+C$\rightarrow$ Where $u$ is a function of $x$

$\textcolor{red}{\text{Proof:}}$

$\int \frac{\mathrm{du}}{1 + {u}^{2}}$

$u = \tan \theta$$\rightarrow$$\mathrm{du} = {\sec}^{2} \theta d \left(\theta\right)$

int(du)/(1+u^2)=int(sec^2thetad(theta))/(1+tan^2theta

color(green)(sec^2theta=1+tan^2theta

$\int \frac{{\sec}^{2} \theta d \left(\theta\right)}{1 + {\tan}^{2} \theta} = \int \frac{\left(\cancel{1 + {\tan}^{2} \theta}\right) d \left(\theta\right)}{\cancel{1 + {\tan}^{2} \theta}}$

$= \int d \left(\theta\right) = \theta$

Reverse the Substitution

$u = \tan \theta$color(red)(rarr$\theta = {\tan}^{-} 1 u$

$\therefore \int \frac{\mathrm{du}}{1 + {u}^{2}} = {\tan}^{-} 1 u + C$

Simply by Substituting in this relation

$\int \frac{\mathrm{dx}}{1 + {x}^{2}} = {\tan}^{-} 1 x + C$

Jun 28, 2018

$\arctan x + C$.

#### Explanation:

It is one of the Standard Integral : $\int \frac{1}{1 + {x}^{2}} = \arctan x + C$.

Aliter :

Let, $I = \int \frac{1}{1 + {x}^{2}} \mathrm{dx}$.

Subst. x=tanu. :. dx=sec^2udu, &, u=arctanx.

$\therefore I = \int \frac{1}{1 + {\tan}^{2} u} {\sec}^{2} u \mathrm{du}$,

$= \int \frac{1}{\sec} ^ 2 u {\sec}^{2} u \mathrm{du}$,

$= \int 1 \mathrm{du}$,

$= u$.

$\Rightarrow I = \arctan x + C$.