Integrating ln and arctan by parts

Idea. Differentiating $\ln(1+x^2)$ or $\arctan x$ produces *rational* functions, which we know how to integrate.

First integral. With $u=\ln(1+x^2)$, $v'=1$:
$$\int_0^1\ln(1+x^2)\,dx=\bigl[x\ln(1+x^2)\bigr]_0^1-\int_0^1\frac{2x^2}{1+x^2}\,dx=\ln2-2\int_0^1\left(1-\frac1{1+x^2}\right)dx.$$
The last integral is $1-\frac\pi4$, so the result is
$$\ln2-2+\frac\pi2\approx0.2639.$$

Second integral. With $u=\arctan x$, $v'=1$:
$$\int_0^1\arctan x\,dx=\bigl[x\arctan x\bigr]_0^1-\int_0^1\frac x{1+x^2}\,dx=\frac\pi4-\frac12\ln2\approx0.4388.$$

Check. Both are positive and smaller than the value of the integrand at $x=1$ ($\ln2\approx0.69$, $\frac\pi4\approx0.79$), as they should be for increasing integrands.