# Question 93667

##### 1 Answer
Jan 30, 2018

$\int x {e}^{4} \mathrm{dx} = \frac{{e}^{4} {x}^{2}}{2} + \text{C}$

#### Explanation:

Given: $\int x {e}^{4} \mathrm{dx}$ (unless you meant $\int x {e}^{4 x} \mathrm{dx}$

In this example, ${e}^{4}$ is a constant which we can pull out of the integral. Therefore we can rewrite the integral as:

${e}^{4} \cdot \int x \mathrm{dx}$

To find $\int x \mathrm{dx}$ use the following princple:

intx^adx=(x^(a+1))/(a+1)+"C";x>0#

So $\int x \mathrm{dx} = \frac{{x}^{1 + 1}}{1 + 1} = {x}^{2} / 2 + \text{C}$

Multiplying ${e}^{4}$ we get

${e}^{4} \cdot {x}^{2} / 2 + \text{C"=(e^4x^2)/2+"C}$