Question

$\int \frac{x-1}{x+e^x}dx$ is equal to

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

Answer 1

The correct option is A ${\log }_e\left(\frac{e^x}{x+e^x}\right)+C$

$\int \frac{(x-1)}{(x+e^x)}dx=\int \frac{(x+e^x)-(1+e^x)}{(x+e^x)}dx=\int 1dx-\int \frac{(1+e^x)}{(x+e^x)}dx$
Let $x+e^x=t\Rightarrow 1+e^x=dt$
$=x-\int \frac{1}{t}dt+c_1=x-{\log }_e(x+e^x)+C={\log }_e\left(\frac{e^x}{x+e^x}\right)+C$