# How do you integrate (1+ x) / (1 + x^2) dx?

Jun 9, 2016

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

#### Explanation:

We have:

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

Split up the numerator and into two different integrals:

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

Notice that the first derivative is just the derivative of the arctangent function, that is, $\int \frac{1}{1 + {x}^{2}} \mathrm{dx} = \arctan \left(x\right) + C$.

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

For the remaining integral, let $u = 1 + {x}^{2}$ and $\mathrm{du} = 2 x \mathrm{dx}$.

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

$= \arctan \left(x\right) + \frac{1}{2} \int \frac{\mathrm{du}}{u}$

This is the natural logarithm integral: $\int \frac{\mathrm{du}}{u} = \ln \left(\left\mid u \right\mid\right) + C$

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

Since $u = 1 + {x}^{2}$:

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