Question

$\int \left\{e^x\left[\frac{\left(x^2+1\right)}{{\left(x+1\right)}^2}\right]\right\}dx=$

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

Answer 1

The correct option is D

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

Explanation for the correct option:

Find the value of $\int \left\{e^x\left[\frac{\left(x^2+1\right)}{{\left(x+1\right)}^2}\right]\right\}dx$:

Consider the given Equation as,

$I=\int \left\{e^x\left[\frac{\left(x^2+1\right)}{{\left(x+1\right)}^2}\right]\right\}dx......\left(1\right)$

Let,

$$\begin{gathered} x+1=t \\ d\left(x+1\right)=dt \\ dx=dt \end{gathered}$$ [differentiating w.r.t $x$]

And

$$\begin{gathered} x+1=t \\ \Rightarrow x=t-1 \end{gathered}$$

Substitute the above values in Equation $\left(1\right)$

$$\begin{gathered} I=\int \left\{e^{t-1}\left[\frac{\left({\left(t-1\right)}^2+1\right)}{{\left(t\right)}^2}\right]\right\}dt \\ I=\frac{1}{e}\int e^t\left(\frac{t^2-2t+2}{t^2}\right)\left[\because e^{-1}=\frac{1}{e}\right] \\ I=\frac{1}{e}\left[\int e^{t}dt-2\int \frac{e^t}{t}dt+2\int \frac{e^t}{t^2}dt\right] \end{gathered}$$

We Know that,

$\int uvdx=u\int vdx-\int \left(\frac{du}{dx}\int vdx\right)dx$

Then $I$ becomes,

$$\begin{gathered} I=\frac{1}{e}\left[e^t-\left[2\frac{1}{t}\int e^{t}dt-\int \left(\frac{d\left(t^{-1}\right)}{dt}\int e^{t}dt\right)dt\right]+2\int \frac{e^t}{t^2}dt\right] \\ =\frac{1}{e}\left[e^t-\left[\frac{2}{t}{e}^t+2\int \frac{1}{t^2}{e}^{t}dt\right]+2\int \frac{e^t}{t^2}dt\right] \\ =\frac{1}{e}\left[e^t-\frac{2}{t}{e}^t\right]+C \\ =e^{t-1}-\frac{2}{t}{e}^{t-1}+C \\ I=e^x-\frac{2e^x}{x+1}+c \end{gathered}$$

$$\begin{gathered} \Rightarrow I=\left\{e^x\left(\frac{x+1-2}{x+1}\right)\right\}+c \\ \Rightarrow I=\left\{e^x\left(\frac{x-1}{x+1}\right)\right\}+c \end{gathered}$$

Hence, the correct answer is Option (D).