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

Question

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

Accepted Answer

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

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QUESTION IMAGE

Question

Explanation:

Step1: Let \( t = e^{x} \), then \( dt=e^{x}dx \) and \( dx=\frac{dt}{t} \)

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

Step3: Integrate the decomposed fractions

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

Answer: