Question

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

• A. $\log \left|\frac{x{e}^x}{x{e}^x+1}\right|+\frac{1}{1+x{e}^x}+c$
• B. $\log \left|\frac{x{e}^x}{x{e}^x+1}\right|-\frac{1}{1+x{e}^x}+c$
• C. $\log \left|\frac{x{e}^x+1}{x{e}^x}\right|+\frac{1}{1+x{e}^x}+c$
• D. None of these

Answer 1

The correct option is A

$\log \left|\frac{x{e}^x}{x{e}^x+1}\right|+\frac{1}{1+x{e}^x}+c$

Evaluating the integral

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

Step 1: Dividing and multiplying $I$ by $e^x$

$\int \frac{e^x(x+1)}{e^{x}x{(1+x{e}^x)}^2}dx....\left(i\right)$

Let $t=x{e}^x...\left(ii\right)$

Differentiating $t$ w.r.t $x$ we get

$$\begin{gathered} \frac{dt}{dx}=x\frac{d}{dx}{e}^x+e^x\frac{d}{dx}\left(x\right) \\ \Rightarrow \frac{dt}{dx}=x{e}^x+e^x \\ \Rightarrow dx=\left(x+1\right)e^{x}dt...\left(iii\right) \end{gathered}$$

Substituting $\left(ii\right)\text{and}\left(iii\right)\text{into}\left(i\right)$

$$\begin{gathered} \Rightarrow \int \frac{1}{t{\left(1+t\right)}^2}dt \\ \Rightarrow \int \frac{1+t-t}{t{\left(1+t\right)}^2}dt\left[\text{adding and subtracting t in numerator}\right] \\ \Rightarrow \int \frac{\left(1+t\right)}{t{\left(1+t\right)}^2}dt-\int \frac{t}{t{\left(1+t\right)}^2}dt \\ \Rightarrow \int \frac{1}{t\left(1+t\right)}dt-\int \frac{1}{{\left(1+t\right)}^2}dt \\ \Rightarrow \int \frac{1+t-t}{t\left(1+t\right)}dt-\int \frac{1}{{\left(1+t\right)}^2}dt\left[\text{again adding and subtracting t in numerator into first integral}\right] \\ \Rightarrow \int \frac{\left(1+t\right)}{t\left(1+t\right)}dt-\int \frac{t}{t\left(1+t\right)}dt-\int \frac{1}{{\left(1+t\right)}^2}dt \\ \Rightarrow \int \frac{1}{t}dt-\int \frac{1}{\left(1+t\right)}dt-\int \frac{1}{{\left(1+t\right)}^2}dt \\ \Rightarrow \ln \left|t\right|-\int \frac{1}{\left(1+t\right)}dt-\int \frac{1}{{\left(1+t\right)}^2}dt+c....\left(iv\right)[c\text{in integrating constant}] \end{gathered}$$

Step 2: Substituting $y=1+t$and $dt=dy$into $\left(iv\right)$.

$$\begin{gathered} \Rightarrow \ln \left|t\right|-\int \frac{1}{y}dy-\int \frac{1}{y^2}dy+c \\ \Rightarrow \ln \left|t\right|-\ln \left|y\right|-\int y^{-2}dy+c \\ \Rightarrow \ln \left|t\right|-\ln \left|y\right|-\frac{y^{-2+1}}{-2+1}+c \\ \Rightarrow \ln \left|t\right|-\ln \left|y\right|+\frac{1}{y}+c \\ \Rightarrow \ln \left|t\right|-\ln \left|(1+t)\right|+\frac{1}{(1+t)}+c[\text{substituting}y=(1+t)] \\ \Rightarrow \ln \left|x{e}^x\right|-\ln \left|(1+x{e}^x)\right|+\frac{1}{(1+x{e}^x)}+c[\text{substituting}t=x{e}^x] \\ \Rightarrow \ln \left|\frac{x{e}^x}{(1+x{e}^x)}\right|+\frac{1}{(1+x{e}^x)}+c\left[\because \ln a-\ln b=\ln \left(\frac{a}{b}\right)\right] \end{gathered}$$

Hence, option A is the correct answer.