Question

The integral $\int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}dx}$ is equal to
(a) $\dfrac{-{{e}^{x}}}{x+1}+C$
(b) $\dfrac{{{e}^{x}}}{x+1}+C$
(c) $\dfrac{x{{e}^{x}}}{x+1}+C$
(d) $\dfrac{-x{{e}^{x}}}{x+1}+C$
(e) $\dfrac{{{e}^{x}}}{{{(x+1)}^{2}}}+C$

Answer 1

Hint: We will rearrange the given function in order to simplify it. Adding and subtracting the same number in the numerator will not change the original function. We will use this fact to simplify the given function. We will integrate the simplified function using the method of integration by parts. The formula for integration by parts for functions $u$ and $v$ with respect to $x$ is as follows,
$\int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\cdot \int{vdx} \right)}dx$ .

Complete step-by-step answer:
The function that we have to integrate is $\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}$ . Now, we will add and subtract 1 to $x$ in the numerator, since it will not affect the original function. So, our function will look like $\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}=\dfrac{(x+1-1){{e}^{x}}}{{{(1+x)}^{2}}}$ . Next, we will split our function as follows,
$\dfrac{(x+1-1){{e}^{x}}}{{{(1+x)}^{2}}}=\dfrac{(x+1){{e}^{x}}}{{{(1+x)}^{2}}}-\dfrac{{{e}^{x}}}{{{(1+x)}^{2}}}=\dfrac{{{e}^{x}}}{(1+x)}-\dfrac{{{e}^{x}}}{{{(1+x)}^{2}}}$ .
We will now integrate this simplified version of the function. The function is integrated in the following manner,
$\begin{align}
  & \int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}dx}=\int{\left[ \dfrac{{{e}^{x}}}{(1+x)}-\dfrac{{{e}^{x}}}{{{(1+x)}^{2}}} \right]}dx \\
 & =\int{\dfrac{{{e}^{x}}}{(1+x)}dx-\int{\dfrac{{{e}^{x}}}{{{(1+x)}^{2}}}}dx}
\end{align}$
Now, let us consider the first term of the above equation and integrate it. As it involves two types of functions, the exponential function and an algebraic function, we will have to use the method of integration by parts. The formula for integration by parts for functions $u$ and $v$ with respect to $x$ is given as follows,
$\int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\cdot \int{vdx} \right)}dx$ .
Substituting $u=\dfrac{1}{1+x}$ and $v={{e}^{x}}$ in the above formula, we get
$\int{\dfrac{{{e}^{x}}}{(1+x)}dx=\dfrac{1}{1+x}\int{{{e}^{x}}dx}-\int{\left[ \dfrac{d}{dx}\left( \dfrac{1}{1+x} \right)\int{{{e}^{x}}dx} \right]dx}}$
Simplifying this equation, we have the following,
\[\int{\dfrac{{{e}^{x}}}{(1+x)}dx=\dfrac{1}{1+x}{{e}^{x}}-\int{\left[ -\dfrac{1}{{{(1+x)}^{2}}}{{e}^{x}} \right]dx}}=\dfrac{{{e}^{x}}}{1+x}+\int{\dfrac{{{e}^{x}}}{{{(1+x)}^{2}}}dx}\]
Now, we will substitute the above value in place of the integration of the first term in the integration of the original function. Therefore we get, $\int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}dx}=\dfrac{{{e}^{x}}}{1+x}+\int{\dfrac{{{e}^{x}}}{{{(1+x)}^{2}}}dx-\int{\dfrac{{{e}^{x}}}{{{(1+x)}^{2}}}}dx}$.
Cancelling out the terms with integrals, we get the answer as $\int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}dx}=\dfrac{{{e}^{x}}}{1+x}+C$ where $C$ is the integration constant.

So, the correct answer is “Option (b)”.

Note: We need to understand by observing the function how it should be simplified. The simplification should help us realize which method of integration needs to be used. Application of the formula for integration by parts needs to be done carefully since the second term in it has an integral inside an integral.