# How do you integrate int ln(1+x^2) by parts from [0,1]?

Feb 14, 2017

${\int}_{0}^{1} \ln \left(1 + {x}^{2}\right) \mathrm{dx} = \ln 2 - 2 + \frac{\pi}{2}$

#### Explanation:

We can take $x$ as finite part:

${\int}_{0}^{1} \ln \left(1 + {x}^{2}\right) \mathrm{dx} = {\left[x \ln \left(1 + {x}^{2}\right)\right]}_{0}^{1} - {\int}_{0}^{1} x d \left(\ln \left(1 + {x}^{2}\right)\right)$

$\left(1\right) {\int}_{0}^{1} \ln \left(1 + {x}^{2}\right) \mathrm{dx} = \ln 2 - 2 {\int}_{0}^{1} \frac{{x}^{2} \mathrm{dx}}{1 + {x}^{2}}$

Now solving the resulting integral:

${\int}_{0}^{1} \frac{{x}^{2} \mathrm{dx}}{1 + {x}^{2}} = {\int}_{0}^{1} \frac{{x}^{2} + 1 - 1}{1 + {x}^{2}} \mathrm{dx}$

${\int}_{0}^{1} \frac{{x}^{2} \mathrm{dx}}{1 + {x}^{2}} = {\int}_{0}^{1} \left(1 - \frac{1}{1 + {x}^{2}}\right) \mathrm{dx}$

${\int}_{0}^{1} \frac{{x}^{2} \mathrm{dx}}{1 + {x}^{2}} = {\int}_{0}^{1} \mathrm{dx} - {\int}_{0}^{1} \frac{\mathrm{dx}}{1 + {x}^{2}}$

${\int}_{0}^{1} \frac{{x}^{2} \mathrm{dx}}{1 + {x}^{2}} = {\left[x\right]}_{0}^{1} - {\left[\arctan x\right]}_{0}^{1}$

${\int}_{0}^{1} \frac{{x}^{2} \mathrm{dx}}{1 + {x}^{2}} = 1 - \frac{\pi}{4}$

Substitute in (1):

${\int}_{0}^{1} \ln \left(1 + {x}^{2}\right) \mathrm{dx} = \ln 2 - 2 + \frac{\pi}{2}$