Question

Find $\int {\left\{ {\dfrac{{\left( {x + 1} \right)}}{{\left[ {x{{\left( {1 + x{e^x}} \right)}^2}} \right]}}} \right\}} dx = $
$\eqalign{
  & 1)\log \left| {\left( {\dfrac{{x{e^x}}}{{x{e^x} + 1}}} \right)} \right| + \dfrac{1}{{\left( {1 + x{e^x}} \right)}} + c \cr
  & 2)\log \left| {\left( {\dfrac{{x{e^x}}}{{x{e^x} + 1}}} \right)} \right| - \dfrac{1}{{\left( {1 + x{e^x}} \right)}} + c \cr
  & 3)\log \left| {\left( {\dfrac{{x{e^x} + 1}}{{x{e^x}}}} \right)} \right| + \dfrac{1}{{\left( {1 + x{e^x}} \right)}} + c \cr} $
4) None of these

Answer 1

Hint: The expression contains an exponential function. We see from the above question that there is no formula relating to the given function. Therefore, we use the substitution method to solve the given problem. The substitution we will be making will be for the function, $x{e^x}$. Then we need to differentiate it and substitute those values to get the Integral value of the given expression.
The formulas we will be using to solve the above problem are:
$\eqalign{
  & \dfrac{{d{x^n}}}{{dx}} = n{x^{(n - 1)}} \cr
  & \int {\log x.dx = \dfrac{1}{x}} + c \cr} $

Complete step-by-step answer:
Let, $I = \int {\left\{ {\dfrac{{\left( {x + 1} \right)}}{{\left[ {x{{\left( {1 + x{e^x}} \right)}^2}} \right]}}} \right\}} dx$
Multiply and divide the expression by ${e^x}$
We get, $I = \int {\left\{ {\dfrac{{{e^x}\left( {x + 1} \right)}}{{\left[ {x{e^x}{{\left( {1 + x{e^x}} \right)}^2}} \right]}}} \right\}} dx$
Now, we make a substitution to simplify the denominator
Let us, put $t = x{e^x}$
Now, by differentiating on both sides, we get
${e^x}(x + 1)dx = dt$
Substituting the above values in the given problem, we get
$I = \int {\dfrac{{dt}}{{t{{(t + 1)}^2}}}} $
Simplifying the terms, we get
$I = \int {\dfrac{{t + 1 - t}}{{t(t + 1)}}} dt$
Splitting the terms,
$I = \int {\dfrac{1}{t}} dt - \int {\dfrac{1}{{t + 1}}} dt$
Using the formula,
$I = \log t - \log (t + 1) + c$
Taking out $\log $common,
$I = \log \left( {\dfrac{t}{{t + 1}}} \right) + c$
Now, we shall put back the value of $t$ in the above equation,
$ \Rightarrow I = \log \left| {\left( {\dfrac{{x{e^x}}}{{1 + x{e^x}}}} \right)} \right| + c$
Therefore, the final answer is $\log \left| {\left( {\dfrac{{x{e^x}}}{{1 + x{e^x}}}} \right)} \right| + c$
Hence, option (4) is the correct answer.
So, the correct answer is “Option 4”.

Note: The given question is very simple and asks us to find the integral value of the given expression. Since there is no particular formula involving those terms, use the substitution method only. Memorize the given formula since they will be used in the problem. And, note that the problem involves both differentiation and integration. The question involves an exponential function, but we do not substitute the formula related to the exponential function.