Question

1. - / 1 points details my notes

make a substitution to express the integrand as a rational function and then evaluate the integral. (remember the constant of integration.)
$$ int \frac { d x } { 5 + e ^ { x } } $$

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:

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