Question

Вычислить неопределенный интеграл

Answer 1

Ответ:

решение на фотографиях

Answer 2

$1); ; int dfrac{e^{x}+sinx}{(e^{x}-cosx)^2}, dx=Big[; u=e^{x}-cosx; ,; du=(e^{x}+sinx)dx; Big]=\\=int dfrac{du}{u^2}=-dfrac{1}{u}+C=-dfrac{1}{e^{x}-cosx}+C=dfrac{1}{cosx-e^{x}}+C$

$2); ; int dfrac{cosx, dx}{sqrt{2+cos2x}}=Big[; cos2x=1-2sin^2x; Big]=int dfrac{cosx, dx}{sqrt{3-2sin^2x}}=\\=Big[; u=sqrt2sinx; ,; du=sqrt2, sinx, dx; Big]=dfrac{1}{sqrt2}int dfrac{du}{sqrt{3-u^2}}=\\=dfrac{1}{sqrt2}cdot arcsinfrac{u}{sqrt3}+C=dfrac{1}{sqrt2}cdot arcsinfrac{sqrt2, sinx}{sqrt3}+C$

$3); ; int dfrac{dx}{xsqrt{1+x^2}}=Big[; x=frac{1}{t}; ,; dx=-frac{dt}{t^2}; Big]=int dfrac{-frac{dt}{t^2}}{frac{1}{t}sqrt{1+frac{1}{t^2}}}=-int dfrac{dt}{t, sqrt{frac{1+t^2}{t^2}}}=\\=-int frac{dt}{sqrt{1+t^2}}=-lnBig|t+sqrt{1+t^2}Big|+C=\\=-lnBig|frac{1}{x}+frac{sqrt{1+x^2}}{x}Big|+C=-lnBig|frac{1+sqrt{1+x^2}}{x}Big|+C$

$4); ; int dfrac{2^{x}, dx}{sqrt{1-4^{x}}}=dfrac{1}{ln2} int dfrac{d(2^{x})}{sqrt{1-(2^{x})^2}}=dfrac{1}{ln2}cdot arcsin(2^{x})+C$

$5); ; int sinsqrt{x}, dx=Big[; x=t^2; ,; dx=2t, dt; ,; sqrt{x}=t; Big]=int sintcdot 2t, dt=\\=Big[; u=t; ,; du=dt; ,; v=int sint, dt=-cost; Big]=2cdot Big(-tcdot cost+int cost, dtBig)=\\=-2t, cost+2, sint+C=-2sqrt{x}cdot cossqrt{x}+2, sinsqrt{x}+C$

$6); ; int dfrac{dx}{sqrt{1+e^{2x}}}=Big[; 1+e^{2x}=t^2; ,; e^{2x}=t^2-1; ,; 2x=ln(t^2-1); ,\\x=frac{1}{2}ln(t^2-1); ,; dx=dfrac{2t, dt}{2(t^2-1)}; Big]=int dfrac{t, dt}{(t^2-1)cdot t}=int dfrac{dt}{t^2-1}=\\\=frac{1}{2}cdot lnBig| dfrac{t-1}{t+1}Big|+C=dfrac{1}{2}cdot lnBig|dfrac{sqrt{1+e^{2x}}-1}{sqrt{1+e^{2x}}+1}Big|+C$

Answer 3

прошу вас помогите пожалуйста