Question

The value of $\int_{0}^{1}{\dfrac{x{{e}^{x}}}{{{\left( 1+x \right)}^{2}}}dx}$ is equal to:

Answer 1

Hint: Write $x$ in the form of $(x+1)-1$ in the numerator. Multiply ${{e}^{x}}$ with these two terms and separate the integral into two parts. Assume the overall integral as $I$ and use ILATE to solve the first part of the integral, while leaving the second part as it is. It will automatically get cancelled due to the simplification of the first part.

Complete step-by-step answer:
We already know about integration but here we are going to use a method called integration by parts. Let us know about this method. In calculus, and more generally in mathematical analysis, integration by parts or partial integration is a process that finds the integral of a product of functions in terms of the integral of the product of their derivative and anti-derivative. It is frequently used to transform the anti-derivative of a product of functions into an anti-derivative for which a solution can be found more easily.
If $u=u(x)$ and $du=u'(x)dx$ while $v=v(x)$ and $dv=v'(x)dx$, then the integration by parts formula states that: $\int{uvdx}=u\int{vdx}-\int{\left[ \dfrac{du}{dx}\times \left( \int{vdx} \right) \right]dx}$. What we have done is, in the first part $u$ is multiplied to the integral of $v$ and in the second part we have taken the integration of product of differentiation of $u$ and integration of v. Now, the problem is which function is to be considered as $u$ and which function as $v$. So, there is a rule called ILATE rule for this.
I- Inverse function
L- Logarithmic function
A- Algebraic function
T- Trigonometric function
E- Exponential function
If there is product of any of these two functions then consider then consider ‘$u$’ as the function which comes first according to ILATE. For example, in a product of inverse and logarithmic function, consider inverse function as ‘$u$’.
Now, let us come to the question. We have,
$I=\int_{0}^{1}{\dfrac{x{{e}^{x}}}{{{\left( 1+x \right)}^{2}}}dx}$
This can be written as,
$I=\int_{0}^{1}{\dfrac{\\{(x+1)-1\\}{{e}^{x}}}{{{\left( 1+x \right)}^{2}}}dx}$
Separating the integral into two parts, we get,
$\begin{aligned} & I=\int_{0}^{1}{\dfrac{(x+1){{e}^{x}}}{{{\left( 1+x \right)}^{2}}}dx}-\int_{0}^{1}{\dfrac{{{e}^{x}}}{{{\left( 1+x \right)}^{2}}}dx} \\\ & =\int_{0}^{1}{\dfrac{{{e}^{x}}}{\left( 1+x \right)}dx}-\int_{0}^{1}{\dfrac{{{e}^{x}}}{{{\left( 1+x \right)}^{2}}}dx} \\\ \end{aligned}$
Now, using ILATE in the first part by assuming, $\dfrac{1}{(1+x)}\text{ as }u\text{ and }{{\text{e}}^{x}}\text{ as }v$, we get,
$\begin{aligned} & I=\left[ \dfrac{{{e}^{x}}}{(1+x)} \right]_{0}^{1}-\int_{0}^{1}{\dfrac{-{{e}^{x}}}{{{\left( 1+x \right)}^{2}}}dx}-\int_{0}^{1}{\dfrac{{{e}^{x}}}{{{\left( 1+x \right)}^{2}}}dx} \\\ & =\left[ \dfrac{{{e}^{x}}}{(1+x)} \right]_{0}^{1}+\int_{0}^{1}{\dfrac{{{e}^{x}}}{{{\left( 1+x \right)}^{2}}}dx}-\int_{0}^{1}{\dfrac{{{e}^{x}}}{{{\left( 1+x \right)}^{2}}}dx} \\\ & =\left[ \dfrac{{{e}^{x}}}{(1+x)} \right]_{0}^{1} \\\ \end{aligned}$
Substituting the limits, we get,
$\begin{aligned} & I=\dfrac{{{e}^{1}}}{1+1}-\dfrac{{{e}^{0}}}{1+0} \\\ & =\dfrac{e}{2}-\dfrac{1}{1} \\\ & =\dfrac{e}{2}-1 \\\ \end{aligned}$

Note: We can see that we have left the integration of $\int_{0}^{1}{\dfrac{{{e}^{x}}}{{{\left( 1+x \right)}^{2}}}dx}$, which is the second part of the integral $I$. This is because if we will try to integrate this, then it will form an infinite series which will never end. Thus, we have left this part to get cancelled due to the simplification of the first part.