Question

помогите пожалуйстаа)))

Answer 1

$intlimits_1^5 {frac{1}{x+sqrt{2x-1}}} , dx = left|begin{array}{c}sqrt{2x-1}=t, \ x=frac{t^2+1}{2}, \ dx=tdt, \ t_1=1, t_2=3. end{array}right| = begin intlimits_1^3 {frac{2t}{t^2+2t+1} , dt end. = \ = begin intlimits_1^3 {frac{2(t+1)}{(t+1)^2} , dt - intlimits_1^3 {frac{2}{(t+1)^2} , dt = 2intlimits_1^3 {frac{1}{t+1} , d(t+1) - 2intlimits_1^3 {frac{1}{(t+1)^2} , d(t+1)end. =$
$= left 2(ln|t+1|+frac{1}{t+1}) right|_1^3 = 2( ln|3+1|+frac{1}{3+1}-ln|1+1|-frac{1}{1+1}) = \ = 2(2ln2+frac{1}{4}-ln2-frac{1}{2}) = 2ln2-frac{1}{2}$