計算：\(\int_{0}^{1}\frac{dx}{(x+1)\sqrt{x^2+1}}\)

試試散裝復變

\[\int_{0}^{1}\frac{dx}{(x+1)\sqrt{x^2+1}} \\ =\int_{0}^{\frac{\pi}{4}}\frac{\sec^2tdt}{(1+\tan t)\sec t} \\ =\int_{0}^{\frac{\pi}{4}}\frac{dt}{\sin t+\cos t} \\ =\int_{0}^{\frac{\pi}{4}}\frac{dt}{\frac{e^{it}-e^{-it}}{2i}+\frac{e^{it}+e^{-it}}{2}} \\ =2\int_{0}^{\frac{\pi}{4}}\frac{dt}{-ie^{it}+ie^{-it}+e^{it}+e^{-it}} \\ =2\int \frac{1}{i}\frac{de^{it}}{-ie^{2it}+i+e^{2it}+1} \\ \xlongequal{u=e^{it}}\frac{-2i}{1-i}\oint_{L(t)} \frac{du}{u^2+\frac{1+i}{1-i}} \\ =\frac{-2i}{(1-i)\sqrt{\frac{1+i}{1-i}}}\oint_{L(t)} \frac{d\frac{u}{\sqrt{\frac{1+i}{1-i}}}}{\left(\frac{u}{\sqrt{\frac{1+i}{1-i}}}\right)^2+1} \\ =\frac{-2i}{(1-i)\sqrt{\frac{1+i}{1-i}}} \arctan \frac{e^{it}}{\sqrt{\frac{1+i}{1-i}}} \bigg|_{L(t)} \] \[\frac{e^{it}}{\sqrt{\frac{1+i}{1-i}}}=\frac{e^{it}}{\sqrt{ \frac{\sqrt{2}e^{i\frac{\pi}{4}}}{\sqrt{2}e^{i(-\frac{\pi}{4})}} }} =\frac{e^{it}}{\sqrt{e^{\frac{i\pi}{2}}}} =e^{i(t-\frac{\pi}{4})} \] \[RHS=\frac{-\sqrt{2}i}{e^{-\frac{i\pi}{4}}e^{\frac{i\pi}{4}}}\arctan e^{i(t-\frac{\pi}{4})} \bigg|_{L(t)} =-\sqrt{2}i\arctan e^{it} \bigg|_{-\frac{\pi}{4} \le t \le 0}=-\sqrt{2}{i}\left(-\frac{i}{2}\right) \mathrm{Ln}(\frac{1+ie^{it}}{1-ie^{it}}) \bigg|_{-\frac{\pi}{4} \le t \le 0} \\ =-\frac{\sqrt{2}}{2} \mathrm{Ln}\left( \frac{1+ie^{it}}{1-ie^{it}} \right) \bigg|_{R(t)} \\ =-\frac{\sqrt{2}}{2}(\ln \vert z \vert + i \arg z) \bigg|_{R(t)} \]\[\frac{1+ie^{it}}{1-ie^{it}}=\frac{1+i(\cos t+i\sin t)}{1-i(\cos t+i\sin t)}=\frac{1-\sin t+i\cos t}{1+\sin t-i\cos t}=\frac{(1-\sin t+i\cos t)(1+\sin t+i\cos t)}{(1+\sin t)^2+\cos^2 t} \\ =\frac{ 1-\sin^2t-\cos^2 t +2i\cos t}{(1+\sin t)^2+\cos^2 t}=\frac{2\cos t}{(1+\sin t)^2+\cos^2 t}i \\ =\frac{\cos t}{1+\sin t}i \]\[RHS=-\frac{\sqrt{2}}{2} \left(\ln \left\vert \frac{\cos t}{1+ \sin t} \right\vert+i \frac{\pi}{2} + 2k\pi i \right) \bigg|_{-\frac{\pi}{4} \le x \le 0} \\ =-\frac{\sqrt{2}}{2}\left(i \frac{\pi}{2}+2k\pi i\right)+\frac{\sqrt{2}}{2} \left( \ln(\sqrt{2}+1)+i \frac{\pi}{2}+2k'\pi i \right) \\ =\frac{\sqrt{2}}{2}\ln(\sqrt{2}+1)+i\sqrt{2}k''\pi \]\[\mathrm{Re}(RHS)=\frac{\sqrt{2}}{2}\ln(\sqrt{2}+1) \]

綜上，\(\int_{0}^{1}\frac{dx}{(x+1)\sqrt{1+x^2}}=\frac{\sqrt{2}}{2}\ln(1+\sqrt{2})\)