Question

помогите пожалуйста, с пояснением если можете плз

Answer 1

$1); ; int _0^4 frac{dx}{1+sqrt{2x+1}} =[; t^2=2x+1; ,; t=sqrt{2x+1}; ,; x=frac{1}{2}(t^2-1); ,\\dx=t, dt; ,; ; t_1=1; ,; t_2=3; ]=int _1^3frac{t, dt}{1+t}=int _1^3(1-frac{1}{1+t})dt=\\=(t-ln|1+t|)|_1^3=3-ln4-(1-ln2)=2+ln2-ln4=\\=2+lnfrac{2}{4}=2+lnfrac{1}{2}=2+ln2^{-1}=2-ln2; ;$

$2); ; intlimits^1_0 frac{x^2cdot dx }{sqrt{4-x^2}}=[; x=2sint; ,; dx=2cost, dt; ,; 4-x^2=4-4sin^2t=\\=4(1-sin^2t)=4cos^2t; ,; sint=frac{x}{2}; ,; t=arcsinfrac{x}{2}; ,\\t_1=0,; t_2=arcsinfrac{1}{2}= frac{pi}{6}; ]=intlimits _0^{frac{pi}{6}} frac{4sin^2tcdot 2cost, dt}{sqrt{4cos^2t}} = intlimits^{frac{pi}{6}}_0 frac{4sin^2tcdot 2cost, dt}{2cost} =\\=4cdot int limits _0^{frac{pi}{6}}frac{1}{2}, (1-cos2t)dt=2cdot (t-frac{1}{2}sin2t)|_0^{frac{pi}{6}}=$

$=2cdot (frac{pi}{6}- frac{1}{2} cdot sinfrac{pi}{3})=frac{pi}{3}-frac{sqrt3}{2} \\3); ; intlimits^1_0 frac{dx}{e^{x}+1}=[; t=e^{x}+1; ,; e^{x}=t-1; ,; x=ln(t-1); ,; dx=frac{dt}{t-1}; ,\\t_1=e^0+1=1+1=2; ,; t_2=e+1; ]= intlimits^{e+1}_2 frac{dt}{tcdot (t-1)}=\\= intlimits^{e+1}_2 Big (frac{1}{t-1}-frac{1}{t}Big )dt= Big (ln|t-1|-ln|t|Big )|_2^{e+1}=\\=lne-ln(e+1)-Big (ln1-ln2Big )=lnfrac{e}{e+1}-0+ln2=\\=ln frac{2e}{e+1} ; ;$

$4); ; intlimits^{frac{a}{2}}_0sqrt{ frac{x}{a-x} }dx=[; t^2= frac{x}{a-x} ; ,; t^2(a-x)=x; ,; t^2a-t^2x=x; ,\\t^2a=t^2x+x; ,; t^2a=x(t^2+1); to ; x= frac{t^2a}{t^2+1} ; ,\\dx= frac{2at(t^2+1)-t^2acdot 2t}{(t^2+1)^2} dt= frac{2at, dt}{(t^2+1)^2} ; ,; t_1=0; ,; t_2=1; ]=\\=intlimits^1_0 frac{2acdot t^2, dt}{(t^2+1)^2} =2a intlimits^1_0 frac{tcdot tcdot dt}{(t^2+1)^2} =[; u=t; ,; du=dt; ,; dv=frac{tcdot dt}{(t^2+1)^2}; ,$

$v=frac{1}{2}int frac{2tcdot dt}{(t^2+1)^2} =frac{1}{2}int frac{dz}{z^2}= frac{1}{2} cdot frac{z^{-1}}{-1}=-frac{1}{2(t^2+1)} ; ;; int u, dv=uv-int v, du; ]=\\=2acdot Big ( -frac{t}{2(t^2+1)} |_0^1+ frac{1}{2}cdot intlimits^1_0 frac{dt}{(t^2+1)} Big )=2acdot Big (-frac{1}{4}+ frac{1}{2} cdot arctgt|_0^1Big )=\\=- frac{2a}{4}+ frac{2a}{2} cdot (arctg1-arctg0)=- frac{a}{2} +acdot (frac{pi}{4}-0)=frac{api}{4}-frac{a}{2}= frac{a(pi -2)}{4}; ;$