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# 微积分(A)随缘一题[24]

7天前 4次浏览

[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}} \
=2int_{0}^{frac{pi}{4}}frac{dt}{-ie^{it}+ie^{-it}+e^{it}+e^{-it}} \
=2int 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{dfrac{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^{ifrac{pi}{4}}}{sqrt{2}e^{i(-frac{pi}{4})}} }}
=frac{e^{it}}{sqrt{e^{frac{ipi}{2}}}}
=e^{i(t-frac{pi}{4})}
]

[RHS=frac{-sqrt{2}i}{e^{-frac{ipi}{4}}e^{frac{ipi}{4}}}arctan e^{i(t-frac{pi}{4})} bigg|_{L(t)}
=-sqrt{2}iarctan 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+isin t)}{1-i(cos t+isin t)}=frac{1-sin t+icos t}{1+sin t-icos t}=frac{(1-sin t+icos t)(1+sin t+icos t)}{(1+sin t)^2+cos^2 t} \
=frac{ 1-sin^2t-cos^2 t +2icos t}{(1+sin t)^2+cos^2 t}=frac{2cos t}{(1+sin t)^2+cos^2 t}i \
=frac{cos t}{1+sin t}i
]

[RHS=-frac{sqrt{2}}{2} left(ln leftvert frac{cos t}{1+ sin t} rightvert+i frac{pi}{2} + 2kpi i right) bigg|_{-frac{pi}{4} le x le 0} \
=-frac{sqrt{2}}{2}left(i frac{pi}{2}+2kpi iright)+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)+isqrt{2}k”pi
]

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