Question

How do you integrate by parts $\left( {x \cdot {e^{4x}}} \right)dx$ ?

Answer 1

Hint : This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. We need to know the basic formulae of integral functions. Also, we need to know the process of solving differentiation functions and integration functions. Also, we need to know how to convert the mixed fraction terms into simple fraction terms.

Complete step-by-step answer :
The given question is shown below,
$\int {\left( {x \cdot {e^{4x}}} \right)dx = ? \to \left( 1 \right)}$
We know that,
$\int {u \cdot dv = uv - \int {vdu \to \left( 2 \right)} }$
Let’s compare the equation $\left( 1 \right)$ and $\left( 2 \right)$ ,
$\left( 1 \right) \to \int {\left( {x \cdot {e^{4x}}} \right)dx = ?}$
$\left( 2 \right) \to \int {u \cdot dv = uv - \int {vdu} }$
We get,

$$u = x \\\ dv = {e^{4x}}dx \;$$

To solve the given question we have to find the value of $du$ and $v$ from the above two equations.
Let’s find $du$ ,
Here $u = x$
To find $du$ we would differentiate the above equation. So, we get
$\dfrac{{du}}{{dx}} = 1$
So, we get
$du = dx \to \left( 3 \right)$
Let’s find $v$
Here $dv = {e^{4x}}dx$
To find $v$ we would integrate the above equation. So, we get
$\int {dv} = \int {{e^{4x}}dx}$
So, we get
$v = \dfrac{{{e^{4x}}}}{4} \to \left( 4 \right)$
(Here we use the formula
$\int {{e^{nx}}} = \dfrac{{{e^{nx}}}}{n}$ )
Let’s substitute $u = x,du = dx,v = \dfrac{{{e^{4x}}}}{4}$ and $dv = {e^{4x}}dx$ in the equation $\left( 2 \right)$ , we get
$\left( 2 \right) \to \int {u \cdot dv = uv - \int {vdu} }$
$\int {x \cdot {e^{4x}}} dx = \left( x \right)\left( {\dfrac{{{e^{4x}}}}{4}} \right) - \int {\dfrac{{{e^{4x}}}}{4}} dx$
$\Rightarrow \int {x \cdot {e^{4x}}} dx = \dfrac{{x \cdot {e^{4x}}}}{4} - \left( {\dfrac{{{e^{4x}}}}{{4 \times 4}} + c} \right)$
(Here $\int {\dfrac{{{e^{4x}}}}{4}} = \dfrac{{{e^{4x}}}}{{4 \times 4}}$ was defined by the formula $\int {{e^{nx}}} = \int {\dfrac{{{e^{nx}}}}{n}}$ )
So, we get
$\Rightarrow \int {x \cdot {e^{4x}}} dx = \dfrac{{x \cdot {e^{4x}}}}{4} - \left( {\dfrac{{{e^{4x}}}}{{4 \times 4}} + c} \right)$
Here we have $\dfrac{{{e^{4x}}}}{4}$ is common in both terms in the RHS of the above equation. So, let’s take the mentioned term as a common term. So, we get
$\Rightarrow \int {x \cdot {e^{4x}}} dx = \dfrac{{{e^{4x}}}}{4}\left( {x - \dfrac{1}{4} + c} \right)$
We know that,
$\dfrac{{{e^{4x}}}}{4} \cdot c = c$ , because $c$ is a constant term.
So, we get
$\Rightarrow \int {x \cdot {e^{4x}}} dx = \dfrac{{{e^{4x}}}}{4}\left( {x - \dfrac{1}{4}} \right) + c \to \left( 5 \right)$
We know that,
$a - \dfrac{b}{c} = \dfrac{{ac - b}}{c}$
By using this formula, we get
$x - \dfrac{1}{4} = \dfrac{{4x - 1}}{4}$
Let’s substitute this value in the equation $\left( 5 \right)$ , we get
$\int {x \cdot {e^{4x}}} dx = \dfrac{{{e^{4x}}}}{4}\left( {\dfrac{{4x - 1}}{4}} \right) + c$
Let’s multiply the denominator $4 \times 4 = 16$
So, we get
$\Rightarrow \int {x \cdot {e^{4x}}} dx = \dfrac{{{e^{4x}}}}{{16}}\left( {4x - 1} \right) + c$
So, the final answer is,
$\int {x \cdot {e^{4x}}} dx = \dfrac{1}{{16}}{e^{4x}} \cdot \left( {4x - 1} \right) + c$
So, the correct answer is “ $\int {x \cdot {e^{4x}}} dx = \dfrac{1}{{16}}{e^{4x}} \cdot \left( {4x - 1} \right) + c$ ”.

Note : Remember the basic formulae involved in the integration process. Note that if we want to find $v$ from $dv$ , we would integrate the term $dv$ . If we want to find $dv$ from $v$ , we would differentiate the term $v$ . Also, remember the algebraic formulae to convert the mixed fraction terms into simple fraction terms.