Giải thích các bước giải:
$\sqrt{e^{2x}+1}(2f(x)+f'(x))=1$
$\to (e^{2x}+1)(2f(x)+f'(x))=\sqrt{e^{2x}+1}$
$\to 2f(x).(e^{2x}+1)+f'(x)(e^{2x}+1)=\sqrt{e^{2x}+1}$
$\to \int 2f(x).(e^{2x}+1)+f'(x)(e^{2x}+1)dx=\int \sqrt{e^{2x}+1}dx$
$\to (e^{2x}+1)f(x)=\dfrac{-1}{2}\ln|\sqrt{e^{2x}+1}+1|+\dfrac{1}{2}\ln|\sqrt{e^{2x}+1}-1|+\sqrt{e^{2x}+1}+C$
Vì $f(0)=\sqrt{2}\to C=\sqrt{2} +\dfrac{1}{2}\ln(\sqrt{2}+1)-\dfrac{1}{2}\ln(\sqrt{2}-1)=\sqrt{2}+\ln (3+2\sqrt{2})$
$\to f(x)=\dfrac{\dfrac{-1}{2}\ln|\sqrt{e^{2x}+1}+1|+\dfrac{1}{2}\ln|\sqrt{e^{2x}+1}-1|+\sqrt{e^{2x}+1}+\sqrt{2}+\ln (3+2\sqrt{2})}{e^{2x}+1}$
$\to f(\ln \sqrt{15})$
