Question

5^sqrt3-2xdx помогите решить интеграл

Answer 1

$\displaystyle\int 5^{\displaystyle\sqrt{3-2x}}dx=-\int t*5^tdt=-\frac{t*5^t}{ln5}+\frac{1}{ln5}\int\ 5^tdt=\\=-\frac{t*5^t}{ln5}+\frac{5^t}{ln^25}+C=\frac{5^t}{ln^25}(1-ln5*t)+C=\\=\frac{5^\sqrt{3-2x}}{ln^25}(1-ln5*\sqrt{3-2x})+C\\3-2x=t^2;-2dx=2tdt;dx=-tdt\\\\u=t;du=dt\\dv=5^tdt;v=\frac{5^t}{ln5}$