Question

Нужно вычеслить несобственный интеграл:

Answer 1

$\displaystyle\\\int\limits^{\ln(12)}_{\ln {(5)}} \frac{1}{\sqrt{e^x+4} } \, dx=\{t=e^x+4 \}\Rightarrow \int \frac{1}{(t-4)\sqrt{t} }dt=\{ u=\sqrt{t} \}\Rightarrow \\\\\\\Rightarrow 2\int \frac{1}{u^2-4} du=2\int \frac{1}{(u-2)(u+2)}du=2\int \bigg(\frac{1}{4(u-2)}-\frac{1}{4(u+2)} \bigg)du=\\\\\\=2\bigg(\frac{1}{4}\int \frac{1}{u-2}du-\frac{1}{4}\int \frac{1}{u+2}du\bigg)=2\bigg( \frac{\ln(u-2)}{4}-\frac{\ln(u+2)}{4} \bigg) =\\\\\\$

$\displaystyle\\=\frac{\ln(u-2)}{2}-\frac{\ln(u+2)}{2}=\{u=\sqrt{t} \}\Rightarrow \frac{\ln(\sqrt{t}-2)}{2}-\frac{\ln(\sqrt{t}+2)}{2}=\\\\\\=\{t=e^x+4 \}\Rightarrow \frac{\ln(\sqrt{e^x+4}-2)-\ln(\sqrt{e^x+4}+2)}{2} \bigg |^{\ln(12)}_{\ln(5)}=\\\\\\=\frac{\ln(\sqrt{e^{\ln(12)}+4}-2)-\ln(\sqrt{e^{\ln(12)}+4}+2)}{2}-\frac{\ln(\sqrt{e^{\ln(5)}+4}-2)-\ln(\sqrt{e^{\ln(5)}+4}+2)}{2}=\\\\\\=\frac{\ln(\sqrt{12+4}-2)-\ln(\sqrt{12+4}+2)}{2}-\frac{\ln(\sqrt{5+4}-2)-\ln(\sqrt{5+4}+2)}{2}=\\\\\\$$\displaystyle\\=\frac{\ln(2)-\ln(6)}{2}-\frac{\ln(1)-\ln(5)}{2}=-\frac{\ln(3)}{2}+\frac{\ln(5)}{2}=\frac{\ln\bigg(\dfrac{5}{3}\bigg )}{2}$