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

May 10, 2018

$x + {\tan}^{-} 1 \left(x\right) + C$

#### Explanation:

Given: $\int \frac{2 + {x}^{2}}{1 + {x}^{2}} \setminus \mathrm{dx}$.

We can split as follows:

$= \int \frac{1 + 1 + {x}^{2}}{1 + {x}^{2}} \setminus \mathrm{dx}$

$= \int \frac{1 + {x}^{2}}{1 + {x}^{2}} + \frac{1}{1 + {x}^{2}} \setminus \mathrm{dx}$

$= \int \frac{1 + {x}^{2}}{1 + {x}^{2}} \setminus \mathrm{dx} + \int \frac{1}{1 + {x}^{2}} \setminus \mathrm{dx}$

The first integral is trivial, and the second one is a common integral which comes out to $\arctan \left(x\right)$ or ${\tan}^{-} 1 \left(x\right)$.

$= \int 1 \setminus \mathrm{dx} + {\tan}^{-} 1 \left(x\right) + C$

$= x + {\tan}^{-} 1 \left(x\right) + C$