业余的业余

做道题玩玩。

$$\int_0^1\cfrac {t^2}{1+t^2}{\rm\,d}t=1-\arctan t|_0^1=1-\cfrac \pi 4$$

$\begin{align*}\int_0^1\int_0^1 \ln\left(x^\overset{\,}{2}+y^2\right){\rm\,d}x{\rm\,d}y&=\int_0^1 x \ln\left(x^\overset{\,}{2}+y^2\middle)\right|_0^1-\int_0^1 \cfrac {2x^2}{x^2+y^2}{\rm\,d}x\hspace{0.2cm}{\rm\,d}y\\
&=\int_0^1  \ln\left(1+y^2\right)-2+2y\arctan\left(\cfrac 1y\right){\rm\,d}y\\
&=y\ln\left(1+y^2\middle)\right|_0^1-\int_0^1\cfrac{2y^2}{1+y^2}{\rm\,d}y-2+\int_0^1 \arctan\left(\cfrac 1y\right){\rm\,d}y^2\\
&=\ln2-4+\cfrac \pi 2+y^2\arctan\left(\cfrac 1y\middle)\right|_0^1-\int_0^1\cfrac{y^2\cdot(-\cfrac 1{y^2})}{1+\cfrac 1{y^2}}{\rm\,d}y\\
&=\ln2-4+\cfrac \pi{2}+\cfrac \pi 4+\int_0^1\cfrac{y^2}{1+y^2}{\rm\,d}y\\
&=\ln2-4+\cfrac \pi{2}+\cfrac \pi 4+1-\cfrac \pi 4\\
&=\ln 2-3+\cfrac \pi 2
\end{align*}$

业余的业余

$$\int_0^1\int_0^1 \left|\ln\left(x^\overset{\,}{2}+y^2\right)\right|{\rm\,d}x{\rm\,d}y=-\int_0^1\int_0^{\sqrt{1-y^2}} \ln\left(x^\overset{\,}{2}+y^2\right){\rm\,d}x{\rm\,d}y+\int_0^1\int_{\sqrt{1-y^2}}^1 \ln\left(x^\overset{\,}{2}+y^2\right){\rm\,d}x{\rm\,d}y$$