Question

Evaluate :$\int \frac{x{e}^x}{{(1+x)}^2}dx$

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

Answer 1

The correct option is B

$\left[\frac{e^x}{(x+1)}\right]+c$

Evaluating the integral

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

$$\begin{gathered} \Rightarrow \int \frac{(x+1-1)e^x}{{(1+x)}^2}dx \\ \Rightarrow \int \left(\frac{(x+1)e^x}{{(1+x)}^2}-\frac{e^x}{{(1+x)}^2}\right)dx \\ \Rightarrow \int \left(\frac{e^x}{(1+x)}-\frac{e^x}{{(1+x)}^2}\right)dx.......\left(i\right) \end{gathered}$$

Substituting $(1+x)=t$ then $x=t-1$

differentiating it w.r.t $x$ we get $dx=dt$

$$\begin{gathered} \Rightarrow \int \left(\frac{e^{t-1}}{t}-\frac{e^{t-1}}{t^2}\right)dt[from(i)] \\ \Rightarrow \frac{1}{e}\int \left(e^t{t}^{-1}-e^t{t}^{-2}\right)dt \\ \Rightarrow \frac{1}{e}\int e^t{t}^{-1}dt-\int e^t{t}^{-2}dt....\left(ii\right) \end{gathered}$$

Applying the formula of integration by parts into $\left(ii\right)$

$\int f\left(x\right).g\left(x\right).dx=f\left(x\right).\int g\left(x\right).dx-\int (f\text{'}(x).\int g(x).dx).dx$

$$\begin{gathered} \Rightarrow \frac{1}{e}\left[\left(t^{-1}\int e^{t}dt-\int \left(\frac{d}{dt}\left(t^{-1}\right)\int e^{t}dt\right)dt\right)-\int t^{-2}{e}^{t}dt\right] \\ \Rightarrow \frac{1}{e}\left[\left(t^{-1}{e}^t+\int t^{-2}{e}^{t}dt\right)-\int t^{-2}{e}^{t}dt\right]+c \\ \Rightarrow \frac{1}{e}\left(t^{-1}{e}^t\right)+c[c=\mathrm{co}ns\tan t] \\ \Rightarrow \frac{1}{e}\left[{(1+x)}^{-1}{e}^{1+x}\right]+c[substitutingbackt=x+1] \\ \Rightarrow \frac{e^x}{(1+x)}+c \end{gathered}$$

Hence option B is the correct answer.