Question

Найти неопределенные интегралы

Answer 1

$\LARGE \\ 1)\int{e^{3x}\over \sqrt{e^{6x}-7}}\mathrm{dx}=\begin{vmatrix} e^{3x}=t\\ \mathrm{dx}={\mathrm{dt}\over3t} \end{vmatrix}={1\over3}\int{t\mathrm{dt}\over t\sqrt{t^2-7}}={1\over3}\ln{|e^{3x}+\sqrt{e^{6x}-7}|}+C\\ 2)\int{\sin{2x}\over3+\sin^2{x}}\mathrm{dx}=2\int{\sin{x}\cos{x}\over3+\sin^2{x}}\mathrm{dx}=\begin{vmatrix} \sin{x}=t\\ \mathrm{dx}={\mathrm{dt}\over \cos{x}} \end{vmatrix}=2\int {t\mathrm{dt}\over3+t^2}=\int{\mathrm{d(3+t^2)}\over3+t^2}=\ln{|3+t^2|}+C=\ln{|3+\sin^2{x}|}+C\\ 3)\int(2x-1)\sin{3x}\mathrm{dx}=\begin{vmatrix} u=2x-1,\mathrm{du}=2\mathrm{dx}\\ \mathrm{dv}=\sin{3x}\mathrm{dx}, v=-{\cos{3x}\over3} \end{vmatrix}= -{(2x-1)\cos{3x}\over3}-{-2\over3}\int\cos{3x}\mathrm{dx}={2\over9}\sin{3x}-{(2x-1)\cos{3x}\over3}+C$