text: # How do you integrate by parts (x*e^(4x)) dx?
  
  Aug 29, 2015
  
  $\int \left(x \cdot {e}^{4 x} \mathrm{dx}\right) = \frac{1}{16} \cdot {e}^{4 x} \cdot \left(4 x - 1\right) + c$
  
  #### Explanation:
  
  Remember that the formula that allows you to integrate a function by parts looks like this
  
  $\textcolor{b l u e}{\int \left(u \cdot \mathrm{dv}\right) = u \cdot v - \int \left(v \cdot \mathrm{du}\right)} \text{ }$, where
  
  $u$, $v$ - are two functions of $x$;
  $\mathrm{du}$, $\mathrm{dv}$ - their derivatives.
  
  So, you need to identify $u$ and $\mathrm{dv}$, then calculate $\mathrm{du}$ and $v$.
  
  If you take $u = x$ and $\mathrm{dv} = {e}^{4 x}$, you will have
  
  $u = x \implies \mathrm{du} = \mathrm{dx}$
  
  and
  
  $\mathrm{dv} = {e}^{4 x} \implies v = \int \left({e}^{4 x} \mathrm{dx}\right) = \frac{1}{4} \cdot {e}^{4 x}$
  
  Your target integral will thus be
  
  $\int \left(x \cdot {e}^{4 x} \mathrm{dx}\right) = x \cdot \frac{1}{4} \cdot {e}^{4 x} - \int \left(\frac{1}{4} {e}^{4 x} \cdot \mathrm{dx}\right)$
  
  $\int \left(x \cdot {e}^{4 x} \mathrm{dx}\right) = \frac{1}{4} \cdot x \cdot {e}^{4 x} - \frac{1}{4} \cdot \frac{1}{4} \cdot {e}^{4 x} + c$
  
  $\int \left(x \cdot {e}^{4 x} \mathrm{dx}\right) = \frac{1}{4} {e}^{4 x} \left(x - \frac{1}{4}\right) + c$
  
  $\int \left(x \cdot {e}^{4 x} \mathrm{dx}\right) = \textcolor{g r e e n}{\frac{1}{16} \cdot {e}^{4 x} \cdot \left(4 x - 1\right) + c}$