Question

помогите вычислить определенный интеграл
$intlimits^4_0 {frac{sqrt{x}}{4+x}} , dx$

Answer 1

$intlimits^4_0 {frac{sqrt{x}}{x+4}} , dx =\\ t=sqrt{x}\\ dt=frac{1}{2sqrt{x}}dx\ 2sqrt{x}dt=dx\ dx=2tdt\\ = intlimits^{sqrt{4}}_{sqrt {0}} {frac{t}{t^2+4}} , 2tdt= 2*intlimits^2_0 {frac{t^2}{t^2+4}} , dt =2* intlimits^2_0 {frac{t^2+4-4}{t^2+4}} , dt =\\ =2* intlimits^2_0 {(1-frac{4}{t^2+4})} , dt =2* [intlimits^2_0 {1}, dt -4*intlimits^2_0 {frac{1}{t^2+2^2}} , dt]=\\ =2* [t|^2_0 -4*frac{1}{2}*arctg(frac{t}{2})|^2_0]=\\ =2*[2-2*(arctg(1)-arctg(0))]=$

$=2*[2-2*(frac{pi}{4}-0)]=4-pi$