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

Jul 23, 2015

The Answer is $x + \arctan \left(x\right)$

#### Explanation:

First note that : $\frac{2 + {x}^{2}}{1 + {x}^{2}}$ can be written as $\frac{1 + 1 + {x}^{2}}{1 + {x}^{2}} = \frac{1}{1 + {x}^{2}} + \frac{1 + {x}^{2}}{1 + {x}^{2}} = 1 + \frac{1}{1 + {x}^{2}}$

$\implies \int \frac{2 + {x}^{2}}{1 + {x}^{2}} \mathrm{dx} = \int \left[1 + \frac{1}{1 + {x}^{2}}\right] \mathrm{dx} = \int \left[1\right] \mathrm{dx} + \int \left[\frac{1}{1 + {x}^{2}}\right] \mathrm{dx} = x + \int \left[\frac{1}{1 + {x}^{2}}\right] \mathrm{dx} =$

The derivative of $\arctan \left(x\right)$ is $\frac{1}{1 + {x}^{2}}$.

This implies that the antiderivative of $\frac{1}{1 + {x}^{2}}$ is $\arctan \left(x\right)$

And it's on that basis that we can write : $\int \left[1 + \frac{1}{1 + {x}^{2}}\right] \mathrm{dx} = x + \arctan \left(x\right)$

Hence,

$\int \frac{2 + {x}^{2}}{1 + {x}^{2}} \mathrm{dx} = = \int \left[1 + \frac{1}{1 + {x}^{2}}\right] \mathrm{dx} = x + \arctan \left(x\right) + c$

So the antiderivative of $\frac{2 + {x}^{2}}{1 + {x}^{2}}$ is $\textcolor{b l u e}{x + \arctan \left(x\right)}$

$\text{NB :}$

Do not confuse the $a n t i \mathrm{de} r i v a t i v e$ with the indefinite integral

Antiderivative does not involve a constant. In fact finding the antiderivative doesn't mean intergrate!