Question

неможу рішити інтеграли

Answer 1

$1)quad int frac{sqrt{x}, dx}{x(x+1)}=int }frac{dx}{sqrt{x}(x+1)}=[, x=t^2,; dx=2t, dt, ]=int frac{2t, dt}{t(t^2+1)} =\\=2int frac{dt}{t^2+1}=2cdot arctgt+C=2cdot arctgsqrt{x}+C\\2)quad int frac{dx}{1+tgx} =[, t=tgx,; x=arctgt,; dx= frac{dt}{1+t^2} , ]=int frac{dt}{(1+t^2)(1+t)} =I\\ frac{1}{(1+t^2)(1+t)} = frac{At+B}{1+t^2} + frac{C}{1+t} = frac{(At+B)(1+t)+C(1+t^2)}{(1+t^2)(1+t)} \\1=At+At^2+B+Bt+C+Ct^2\\t^2; |; A+C=0\\t; |; A+B=0\\t^0; |; B+C=1$

$A=-C; ,; A=-B; ,; to ; ; -C=-B; ,; ; C=B\\2B=1; ; to ; ; B=frac{1}{2}=C; ,; ; A=-frac{1}{2}\\I=int (frac{-frac{1}{2}t+frac{1}{2}}{1+t^2} + frac{frac{1}{2}}{1+t} )dt=-frac{1}{2}int frac{t, dt}{1+t^2} + frac{1}{2}int frac{dt}{1+t^2} +frac{1}{2}int frac{dt}{1+t}=\\=-frac{1}{4}int frac{d(1+t^2)}{1+t^2} +frac{1}{2}arctgt+frac{1}{2}ln|1+t|=\\=-frac{1}{4}ln|1+t^2|+frac{1}{2}arctgt+frac{1}{2}ln|1+t|+C=$

$=-frac{1}{4}ln(1+tg^2x)+frac{1}{2}arctg(tgx)+frac{1}{2}ln|1+tgx|+C=$

$=-frac{1}{4}ln(1+tg^2x)+frac{1}{2}x+frac{1}{2}ln|1+tgx|+C$


$3)quad intlimits^1_0 {frac{x^3, dx}{3+x^4}} =frac{1}{4}int _0^1 frac{4x^3, dx}{3+x^4} =frac{1}{4}int _0^1 frac{d(3+x^4)}{3+x^4} =frac{1}{4}ln|3+x^4|, |_0^1=\\=frac{1}{4}(ln4-ln3)=frac{1}{4}lnfrac{4}{3}$

Answer 2

СПАСИБО

Answer 3

выручил