Salut.. am nevoie de puțin ajutor.. cât este integrala din e^x + e^(1-x)

[tex]\int \:e^x+e^{1-x}dx \\ \\ =\int \:e^xdx+\int \:e^{1-x}dx \\ \\ \int \:e^xdx=e^x \\ \\ si : \int \:e^{1-x}dx=-e^{1-x} \\ \\ Prin, urmare : =e^x-e^{1-x}=e^x-e^{1-x}+C[/tex]

[tex]\displaystyle\int (e^x+e^{1-x})dx=\int e^xdx+\int e^{1-x} dx=e^x+\int e^{1-x}dx\\ \text{Rezolvam a doua integrala prin metoda schimbarii de variabile:}\\ 1-x\stackrel{not}{=} t\Rightarrow dt= -dx\\ \text{Prin urmare:}\\ \int e^{1-x}dx=-\int e^{t}dt=-e^t+C= -e^{1-x}+C\\ \text{Deci }\boxed{\displaystyle\int (e^x+e^{1-x})dx=e^x-e^{1-x}+C} [/tex]