積分を解く

$${\int }_0^1\frac{\log (x+1)}{x^2+1}dxを求めよ。$$

$$\begin{array}{rlr}& I={\int }_0^1\frac{\log (x+1)}{x^2+1}dx& \\ f(t)& ={\int }_0^1\frac{\log (tx+1)}{x^2+1}dx& \\ f^{\prime }(t)& ={\int }_0^1\frac{x}{(x^2+1)(tx+1)}dx\\ & =\frac{1}{t^2+1}{\int }_0^1(\frac{t+x}{x^2+1}-\frac{t}{tx+1})dx\\ & =\frac{1}{t^2+1}(\frac{\pi }{4}t+\frac{\log 2}{2}-\log (t+1))\end{array}$$

$$\begin{array}{r}求める値はf(1)-f(0)\end{array}$$

$$\begin{array}{rl}f(1)-f(0)& ={\int }_0^1{f}^{\prime }(t)dt\\ & ={\int }_0^1\frac{1}{t^2+1}(\frac{\pi }{4}t+\frac{\log 2}{2}-\log (t+1))dt\\ & =\frac{\pi }{8}\log 2+\frac{\pi }{8}\log 2-I\end{array}$$

$$\begin{array}{r}f(1)-f(0)がIなので、\end{array}$$

$$\begin{array}{r}2I=\frac{\pi }{4}\log 2\end{array}$$

$$\begin{array}{r}よって、求める値は\frac{\pi }{8}\log 2\end{array}$$

$$\begin{array}{rl}I& ={\int }_0^1\frac{\log (x+1)}{x^2+1}dx\\ & ={\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx\\ & ={\int }_0^{\frac{\pi }{4}}\log (1+\tan (\frac{\pi }{4}-x))dx\\ & ={\int }_0^{\frac{\pi }{4}}\log \left(\frac{2}{1+\tan x}\right)\\ & ={\int }_0^{\frac{\pi }{4}}\log 2dx-{\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx\end{array}$$
$$\begin{array}{rl}2I& =\frac{\pi }{4}\log 2\\ I& =\frac{\pi }{8}\log 2\end{array}$$