$\int \frac{x+1}{x\left(1+x \mathrm{e}^x\right)^2} \mathrm{~d} x=$ (where $C$ is a constant of integration.)

$\int \frac{x+1}{x\left(1+x \mathrm{e}^x\right)^2} \mathrm{~d} x=$ (where $C$ is a constant of integration.)
  1. $\log \left(1+x \mathrm{e}^x\right)+\frac{1}{1+x \mathrm{e}^x}+C$
  2. $\log \left(\frac{x \mathrm{e}^x}{1+x \mathrm{e}^x}\right)+\frac{1}{1+x \mathrm{e}^x}+C$
  3. $\log \left(\frac{x \mathrm{e}^x}{1+x \mathrm{e}^x}\right)+C$
  4. $\log \left(\frac{x \mathrm{e}^x}{1+x \mathrm{e}^x}\right)-\frac{1}{1+x \mathrm{e}^x}+C$

Solution

$\begin{aligned} & \int \frac{(x+1) \mathrm{d} x}{x\left(1+x \mathrm{e}^x\right)^2}=\int \frac{\mathrm{e}^x(x+1) \mathrm{d} x}{x \cdot \mathrm{e}^x\left(1+x \mathrm{e}^x\right)^2} \\ & =\int \frac{\mathrm{d} t}{t(1+t)^2} \quad[\operatorname{let} x \cdot \mathrm{e}=t \Rightarrow \mathrm{e}(x+1) \mathrm{d} x=\mathrm{d} t] \\ & =\int\left\{\frac{1}{t}-\frac{1}{1+t}-\frac{1}{(1+t)^2}\right\} \mathrm{d} t \\ & =\log |t|-\log |1+t|+\frac{1}{1+t}+C \\ & =\log \left(\frac{t}{1+t}\right)+\frac{1}{1+t}+C \\ & =\log \left(\frac{x \cdot \mathrm{e}^x}{1+x \cdot \mathrm{e}^x}\right)+\frac{1}{1+x \cdot \mathrm{e}^x}+C\end{aligned}$

Asked in: MHT CET 2022 (05 Aug Shift 1)

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