微积分(A)随缘一题[24]

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

试试散装复变

\[\int_{0}^{1}\frac{dx}{(x+1)\sqrt{x^2+1}} \\ =\int_{0}^{\frac{\pi}{4}}\frac{\sec^2tdt}{(1+\tan t)\sec t} \\ =\int_{0}^{\frac{\pi}{4}}\frac{dt}{\sin t+\cos t} \\ =\int_{0}^{\frac{\pi}{4}}\frac{dt}{\frac{e^{it}-e^{-it}}{2i}+\frac{e^{it}+e^{-it}}{2}} \\ =2\int_{0}^{\frac{\pi}{4}}\frac{dt}{-ie^{it}+ie^{-it}+e^{it}+e^{-it}} \\ =2\int \frac{1}{i}\frac{de^{it}}{-ie^{2it}+i+e^{2it}+1} \\ \xlongequal{u=e^{it}}\frac{-2i}{1-i}\oint_{L(t)} \frac{du}{u^2+\frac{1+i}{1-i}} \\ =\frac{-2i}{(1-i)\sqrt{\frac{1+i}{1-i}}}\oint_{L(t)} \frac{d\frac{u}{\sqrt{\frac{1+i}{1-i}}}}{\left(\frac{u}{\sqrt{\frac{1+i}{1-i}}}\right)^2+1} \\ =\frac{-2i}{(1-i)\sqrt{\frac{1+i}{1-i}}} \arctan \frac{e^{it}}{\sqrt{\frac{1+i}{1-i}}} \bigg|_{L(t)} \]

\[\frac{e^{it}}{\sqrt{\frac{1+i}{1-i}}}=\frac{e^{it}}{\sqrt{ \frac{\sqrt{2}e^{i\frac{\pi}{4}}}{\sqrt{2}e^{i(-\frac{\pi}{4})}} }} =\frac{e^{it}}{\sqrt{e^{\frac{i\pi}{2}}}} =e^{i(t-\frac{\pi}{4})} \]

\[RHS=\frac{-\sqrt{2}i}{e^{-\frac{i\pi}{4}}e^{\frac{i\pi}{4}}}\arctan e^{i(t-\frac{\pi}{4})} \bigg|_{L(t)} =-\sqrt{2}i\arctan e^{it} \bigg|_{-\frac{\pi}{4} \le t \le 0}=-\sqrt{2}{i}\left(-\frac{i}{2}\right) \mathrm{Ln}(\frac{1+ie^{it}}{1-ie^{it}}) \bigg|_{-\frac{\pi}{4} \le t \le 0} \\ =-\frac{\sqrt{2}}{2} \mathrm{Ln}\left( \frac{1+ie^{it}}{1-ie^{it}} \right) \bigg|_{R(t)} \\ =-\frac{\sqrt{2}}{2}(\ln \vert z \vert + i \arg z) \bigg|_{R(t)} \]

\[\frac{1+ie^{it}}{1-ie^{it}}=\frac{1+i(\cos t+i\sin t)}{1-i(\cos t+i\sin t)}=\frac{1-\sin t+i\cos t}{1+\sin t-i\cos t}=\frac{(1-\sin t+i\cos t)(1+\sin t+i\cos t)}{(1+\sin t)^2+\cos^2 t} \\ =\frac{ 1-\sin^2t-\cos^2 t +2i\cos t}{(1+\sin t)^2+\cos^2 t}=\frac{2\cos t}{(1+\sin t)^2+\cos^2 t}i \\ =\frac{\cos t}{1+\sin t}i \]

\[RHS=-\frac{\sqrt{2}}{2} \left(\ln \left\vert \frac{\cos t}{1+ \sin t} \right\vert+i \frac{\pi}{2} + 2k\pi i \right) \bigg|_{-\frac{\pi}{4} \le x \le 0} \\ =-\frac{\sqrt{2}}{2}\left(i \frac{\pi}{2}+2k\pi i\right)+\frac{\sqrt{2}}{2} \left( \ln(\sqrt{2}+1)+i \frac{\pi}{2}+2k'\pi i \right) \\ =\frac{\sqrt{2}}{2}\ln(\sqrt{2}+1)+i\sqrt{2}k''\pi \]

\[\mathrm{Re}(RHS)=\frac{\sqrt{2}}{2}\ln(\sqrt{2}+1) \]

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

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\[\begin{aligned} &I=\int \frac{dx}{\sin x+\cos x} \\ &J= \int \frac{dx}{\sin x-\cos x} \\ &I+J=-2\int \frac{d\cos x}{1-2\cos^2 x}=- \int \frac{d \cos x}{1+\sqrt{2}\cos x}-\int \frac{d\cos x}{1-\sqrt{2}\cos x}=-\frac{1}{\sqrt{2}}\ln \left| \frac{1+\sqrt{2}\cos x}{1-\sqrt{2}\cos x} \right| \\ &I-J=-2\int \frac{d \sin x}{\sin^2 x-\cos^2 x}=-2\int \frac{d\sin x}{2\sin^2 x-1}=\frac{1}{\sqrt{2}}\ln \left| \frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x} \right| \\ &I=\frac{\sqrt{2}}{4} \ln \left| \frac{1-\sin 2x+\sqrt{2}(\sin x-\cos x)}{1-\sin 2x+\sqrt{2}(\cos x-\sin x)} \right| \end{aligned} \]

\[\int \frac{dx}{\sin x+\cos x} \\ =\int \frac{2dx}{e^{ix}+e^{-ix}+ie^{-ix}-ie^{ix}} \\ \xlongequal {t=e^{ix}}\int \frac{2\frac{dt}{it}}{(1-i)t+(1+i)\frac{1}{t}} \\ =(1-i)\int \frac{dt}{t^2+i} \\ \]