make a substitution to express the integrand as a rational functi\n int \frac{d x}{5+e^{x}}

Explanation:

Step1: Let ( t = e^{x} ), then ( dt=e^{x}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(t + 5)} )

Step2: Decompose the fraction (\frac{1}{t(t + 5)})

Using partial - fraction decomposition: (\frac{1}{t(t + 5)}=\frac{A}{t}+\frac{B}{t + 5}) [ \begin{align*} 1&=A(t + 5)+Bt\ 1&=(A + B)t+5A \end{align*} ] We get ( A=\frac{1}{5}) and ( B=-\frac{1}{5}) So (\frac{1}{t(t + 5)}=\frac{1}{5t}-\frac{1}{5(t + 5)})

Step3: Integrate the decomposed fractions

[ \begin{align*} \int\frac{dt}{t(t + 5)}&=\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 \end{align*} ]

Step4: Substitute back ( t = e^{x} )

[ \begin{align*} \frac{1}{5}\ln|e^{x}|-\frac{1}{5}\ln|e^{x}+ 5|+C&=\frac{1}{5}x-\frac{1}{5}\ln(e^{x}+5)+C \end{align*} ]

Answer:

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