Question

If $\int \frac{(x+1)}{x{(1+x{e}^x)}^2}dx=\log |1-f\left(x\right)|+f\left(x\right)+C$, then $f\left(x\right)=$

• A. $\frac{1}{x+e^x}$
• B. $\frac{1}{1+x{e}^x}$
• C. $\frac{1}{{(1+x{e}^x)}^2}$
• D. $\frac{1}{{(x+e^x)}^2}$

Answer 1

The correct option is B $\frac{1}{1+x{e}^x}$

Put $t=x{e}^x\Rightarrow dt=e^x(1+x)dx$

Thus, $\int \frac{(x+1)}{x{(1+x{e}^x)}^2}dx=\int \frac{e^x(1+x)dx}{x{e}^x{(1+x{e}^x)}^2}$

$=\int \frac{dt}{t{(1+t)}^2}=\int (\frac{1}{t}-\frac{1}{1+t}-\frac{1}{{(1+t)}^2})dt$

$=\log \left|t\right|-\log |1+t|+\frac{1}{1+t}+c$

$=\log \left|\frac{t}{1+t}\right|+\frac{1}{1+t}+c=\log \left|(1-\frac{1}{1+t})\right|+\frac{1}{1+t}+c$

$\therefore f\left(x\right)=\frac{1}{1+x{e}^x}$