# How do you evaluate: indefinite integral (1+x)/(1+x^2) dx?

Mar 24, 2018

The answer is $= \arctan x + \frac{1}{2} \ln \left(1 + {x}^{2}\right) + C$

#### Explanation:

The integral is

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

The first integral

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

And the second integral is

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

$= \frac{1}{2} \ln \left(1 + {x}^{2}\right)$

And finally

$I = \arctan x + \frac{1}{2} \ln \left(1 + {x}^{2}\right) + C$