Question

Интеграл от 3 до а. dx/(3+x)^2=1/30
Найти значение а.

Answer 1

$intlimits^a_3 { frac{1}{(x+3)^2} } , dx = frac{1}{30}$

$intlimits^a_3 {{(x+3)^{-2}} , d(x+3) = frac{1}{30}$

$frac{1}{-2+1}*(x+3)^{-2+1}|_3^a=frac{1}{30}$

$frac{1}{-1}*(x+3)^{-1}|_3^a=frac{1}{30}$

$frac{1}{x+3}|_3^a=-frac{1}{30}$

$frac{1}{a+3}- frac{1}{3+3}=-frac{1}{30}$

$frac{1}{a+3}=frac{1}{6}-frac{1}{30}$

$frac{1}{a+3}=frac{1}{6}(1-frac{1}{5})$

$frac{1}{a+3}=frac{1}{6}*frac{4}{5}$

$frac{1}{a+3}=frac{1}{frac{6*5}{4}}$

$a+3=frac{6*5}{4}$

$a+3=7.5$

$a=4.5$