Question

5^sqrt3-2xdx помогите решить интеграл

Answer 1

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