Question

Evaluate the integral of $\int {{{\text{e}}^{\left( {2{\text{x + log x}}} \right)}}} {\text{dx}} =$
A) $\dfrac{1}{4}\left( {2{\text{x - 1}}} \right){{\text{e}}^{\left( {2{\text{x}}} \right)}} + {\text{c}}$
B) $\dfrac{1}{4}\left( {2{\text{x + 1}}} \right) + \dfrac{2}{{{\text{x + 1}}}} + {\text{c}}$
C) $\dfrac{1}{2}\left( {2{\text{x + 1}}} \right){{\text{e}}^{\left( {2{\text{x}}} \right)}} + {\text{c}}$
D) $\dfrac{1}{2}\left( {2{\text{x - 1}}} \right){{\text{e}}^{\left( {2{\text{x}}} \right)}} + {\text{c}}$

Answer 1

In this type of problem you need to find the value of a given expression by using integration and differentiation methods. The key point in these questions is to find the integral of a given question by using integration and differentiation methods. By using integration and differentiation methods, the given equation has been taken as a ${\text{u , dv}}$ then find the value of ${\text{du , v}}$ . Next, we substitute these values in the integration formula for getting the solution. By integrating the values, we get the value of the given integral.
In this question we have to evaluate the question to find the value of integral of $\int {{{\text{e}}^{\left( {2{\text{x + log x}}} \right)}}} {\text{dx}}$. For that we are going to solve using integration and differentiation methods. And also we are going to substitute the value that has been given in a complete step by step solution.

Formula used: ${\text{log x = }}\dfrac{{\text{1}}}{{\text{x}}}$ where ${\text{x}}$ is a constant.
$\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{uv = u }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ + v }}\dfrac{{{\text{du}}}}{{{\text{dx}}}}$
${\text{u dv = uv - }}\int {{\text{v du}}}$

Complete step-by-step answer:
Here it is given that integral of $\int {{{\text{e}}^{\left( {2{\text{x + log x}}} \right)}}} {\text{dx}}$. We have to find the integral of the given expression.
By using integration and differentiation methods we are going to find the value of given expression.
Now, consider given
$\int {{{\text{e}}^{{\text{2x + log x}}}}{\text{dx = }}\int {\left( {{{\text{e}}^{{\text{2x}}}}{\text{. }}{{\text{e}}^{{\text{log x}}}}} \right)} }$ [ Here we separate the power, while separating the power it convert to multiplication sign)
$\Rightarrow \int {\left( {{{\text{e}}^{{\text{2x}}}}{\text{. x}}} \right)} {\text{ dx}}$ [Here {{\text{\not e}}^{{\text{lo\cancel{g} x}}}} because it is inverse function]
$\Rightarrow \int {\left( {{\text{x }}{\text{. }}{{\text{e}}^{{\text{2x}}}}} \right){\text{ d}}{{\text{x}}^{}}}$
Here let, ${\text{u = x}}$, ${\text{du = 1 dx}}$
${\text{dv = }}{{\text{e}}^{{\text{2x}}}}$ Here we are integrating the terms,
$\int {{\text{dv = }}\int {{{\text{e}}^{2{\text{x}}}}} } \Rightarrow {\text{v = }}\dfrac{{{{\text{e}}^{{\text{2x}}}}}}{{\text{2}}}$
Substitute these values in this formula,
${\text{u dv = uv - }}\int {{\text{v du}}}$
$\int {\left( {{\text{x }}{\text{. }}{{\text{e}}^{{\text{2x}}}}} \right){\text{dx}}} {\text{ = }}\dfrac{{{\text{x}}{{\text{e}}^{{\text{2x}}}}}}{{\text{2}}}{\text{ - }}\int {\dfrac{{{{\text{e}}^{{\text{2x}}}}}}{{\text{2}}}{\text{.1}}} {\text{ dx}}$
Hence we get,
$\Rightarrow \dfrac{{\text{1}}}{{\text{2}}}{\text{x }}{{\text{e}}^{{\text{2x}}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}\int {{{\text{e}}^{{\text{2x}}}}} {\text{dx}}$
Integrating the terms,
$\Rightarrow \dfrac{{\text{1}}}{{\text{2}}}{\text{x }}{{\text{e}}^{{\text{2x}}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}\left( {\dfrac{{{{\text{e}}^{{\text{2x}}}}}}{{\text{2}}}} \right){\text{ + c}}$
Multiplying the terms, we have
$\Rightarrow \dfrac{{\text{1}}}{{\text{2}}}{\text{x}}{{\text{e}}^{{\text{2x}}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{4}}}{{\text{e}}^{{\text{2x}}}}{\text{ + c}}$
Taking $\dfrac{1}{4}$ as common then we have,
$\Rightarrow \dfrac{{\text{1}}}{{\text{4}}}{{\text{e}}^{{\text{2x}}}}\left( {{\text{2x - 1}}} \right){\text{ + c}}$
Hence,
$$\int {{{\text{e}}^{\left( {2{\text{x + log x}}} \right)}}} {\text{dx}} =$$$\dfrac{{\text{1}}}{{\text{4}}}{{\text{e}}^{{\text{2x}}}}\left( {{\text{2x - 1}}} \right){\text{ + c}}$

$\therefore$ Option A is the correct answer.

Note: We use definite integrals when the upper and lower limits of that function are given. We use indefinite integrals when no limits are given to a particular function. Integration is the inverse of differentiation.