9. - / 1 points details my notes\nmake a substitution to express the integrand as a rational function and…

Explanation:

Step1: Substitute ( t = e^{x} )

Since ( t = e^{x} ), then ( dt=e^{x}dx=t dx ), and ( dx=\frac{dt}{t} ). The integral ( \int\frac{dx}{5 + e^{x}} ) becomes ( \int\frac{1}{5 + t}\cdot\frac{dt}{t}=\int\frac{dt}{t(5 + t)} ).

Step2: Decompose the fraction

Use partial - fraction decomposition. Let ( \frac{1}{t(5 + t)}=\frac{A}{t}+\frac{B}{5 + t} ). Then ( 1=A(5 + t)+Bt ). Set ( t = 0 ), we get ( A=\frac{1}{5} ). Set ( t=-5 ), we get ( B =-\frac{1}{5} ). So ( \frac{1}{t(5 + t)}=\frac{1}{5t}-\frac{1}{5(t + 5)} ). The integral ( \int\frac{dt}{t(5 + t)}=\frac{1}{5}\int\frac{dt}{t}-\frac{1}{5}\int\frac{dt}{t + 5} ).

Step3: Integrate each term

(\frac{1}{5}\int\frac{dt}{t}-\frac{1}{5}\int\frac{dt}{t + 5}=\frac{1}{5}\ln|t|-\frac{1}{5}\ln|t + 5|+C). Substitute back ( t = e^{x} ), we have ( \frac{1}{5}\ln(e^{x})-\frac{1}{5}\ln(e^{x}+5)+C). Since ( \ln(e^{x})=x ), the integral is ( \frac{x}{5}-\frac{1}{5}\ln(e^{x}+5)+C ).

Answer:

(\frac{x}{5}-\frac{1}{5}\ln(e^{x}+5)+C)