# How do you find the integral of ln(1+x^2)?

Jun 3, 2018

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

#### Explanation:

Calculate this integral by integration by parts

$\int u v ' = u v - \int u ' v$

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

$v ' = 1$, $\implies$, $v = x$

Therefore, the integral is

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

The second integral is

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

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

$= 2 x - 2 \arctan x$

Finally, the integral is

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