# How do you evaluate the integral int ln(1+x^2)?

Jan 20, 2017

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

#### Explanation:

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

Integration by parts takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. For the given integral, let:

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

$\textcolor{w h i t e}{u = \ln \left(1 + {x}^{2}\right)} \mathrm{du} = \frac{2 x}{1 + {x}^{2}} \mathrm{dx}$

$\mathrm{dv} = \mathrm{dx}$

$\textcolor{w h i t e}{\mathrm{dv} = \mathrm{dx}} v = x$

Then:

$I = u v - \int v \mathrm{du}$

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

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

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

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

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