Question

$\int \frac{1+x}{x+e^{-x}}dx=$

• A. $\log \left|\left(x-e^{-x}\right)\right|+C$
• B. $\log \left|\left(x+e^{-x}\right)\right|+C$
• C. $\log \left|\left(1+x{e}^x\right)\right|+C$
• D. ${\left(1+x{e}^{-x}\right)}^2+C$
• E. $\log \left|\left(1-x{e}^x\right)\right|+C$

Answer 1

The correct option is C

$\log \left|\left(1+x{e}^x\right)\right|+C$

Explanation for Correct answer:

Finding the value for the given integral:

$$\begin{gathered} \int \frac{1+x}{x+e^{-x}}dx=\int \frac{1+x}{x+{\displaystyle \frac{1}{e^x}}}dx \\ =\int \frac{1+x}{{\displaystyle \frac{x{e}^x+1}{e^x}}}dx \\ =\int \frac{e^x\left(1+x\right)}{{\displaystyle x{e}^x+1}}dx \end{gathered}$$

Solving this by using the substitution method

Let $$\begin{gathered} t=x{e}^x+1 \\ dt=\left(x{e}^x+1.e^x\right)dx\left[\because \frac{d}{dx}uv=u.dv+v.du\right] \end{gathered}$$

$$\begin{gathered} \int \frac{1+x}{x+e^{-x}}dx=\int \frac{dt}{t} \\ =\log \left|t\right|+C \\ =\log \left|x{e}^x+1\right|+C \end{gathered}$$

Hence, option (C) is the correct answer.