Question

How do you integrate $ \int{{{x}^{2}}{{e}^{3x}}} $ by integration by parts method?

Answer 1

Hint: We will see the method of integration by parts. We will look at the formula used in this method. Then we will split the given product of functions in the integrand using the ILATE rule and determine which function should be used to compute the derivative. After that we have to repeat these steps to integrate the second term in the formula.

Complete step by step answer:
The method of integration by parts is used when we see a product of two functions in the integrand. If $ u $ and $ v $ are the two functions, then the integration by parts is given as,
\[\int{uvdx=u\int{vdx}-\int{\left( \dfrac{du}{dx}\int{vdx} \right)}}dx\]
We have to integrate the following function using the method of integration by parts,
 $ I=\int{{{x}^{2}}{{e}^{3x}}} $
We have to split the given product of functions in the integrand. We will use the ILATE rule for doing this. The ILATE rule is a rule which tells us the preference order for the type of function that should be substituted in place of function $ u $ in the formula mentioned above. The letters represent the inverse functions, logarithmic functions, algebraic functions, trigonometric functions and exponential functions in order. So, according to this rule we can let $ u={{x}^{2}} $ and $ v={{e}^{3x}} $ .
Therefore, we have $ \dfrac{du}{dx}=\dfrac{d}{dx}\left( {{x}^{2}} \right)=2x $ and $ \int{vdx}=\int{{{e}^{3x}}dx}=\dfrac{{{e}^{3x}}}{3} $ . Substituting these values in the formula mentioned above, we get
 $ \begin{align}
  & I={{x}^{2}}\cdot \dfrac{{{e}^{3x}}}{3}-\int{2x\cdot \dfrac{{{e}^{3x}}}{3}dx} \\
 & \therefore I=\dfrac{{{x}^{2}}{{e}^{3x}}}{3}-\dfrac{2}{3}\int{x{{e}^{3x}}dx} \\
\end{align} $
Let us integrate the second term in the above expression using the same formula. Now, we have $ u=x $ and $ v={{e}^{3x}} $ . So, we have $ \dfrac{du}{dx}=\dfrac{d}{dx}\left( x \right)=1 $ and $ \int{vdx}=\int{{{e}^{3x}}dx}=\dfrac{{{e}^{3x}}}{3} $ .
Using the by parts formula, we get
 $ \begin{align}
  & \int{x{{e}^{3x}}dx}=x\cdot \dfrac{{{e}^{3x}}}{3}-\int{\dfrac{{{e}^{3x}}}{3}dx} \\
 & \Rightarrow \int{x{{e}^{3x}}dx}=\dfrac{x{{e}^{3x}}}{3}-\dfrac{1}{3}\cdot \dfrac{{{e}^{3x}}}{3} \\
 & \therefore \int{x{{e}^{3x}}dx}=\dfrac{x{{e}^{3x}}}{3}-\dfrac{{{e}^{3x}}}{9} \\
\end{align} $
Substituting this value of integration in $ I $ , we get the following,
 $ \begin{align}
  & I=\dfrac{{{x}^{2}}{{e}^{3x}}}{3}-\dfrac{2}{3}\left( \dfrac{x{{e}^{3x}}}{3}-\dfrac{{{e}^{3x}}}{9} \right) \\
 & \Rightarrow I=\dfrac{{{x}^{2}}{{e}^{3x}}}{3}-\dfrac{2x{{e}^{3x}}}{9}+\dfrac{2{{e}^{3x}}}{27} \\
 & \therefore I=\dfrac{{{e}^{3x}}}{3}\left( {{x}^{2}}-\dfrac{2x}{3}+\dfrac{2}{9} \right) \\
\end{align} $

Note:
 There are other methods for integration. These are integration by substitution and integration by partial fractions. It is useful to know all the methods for integration so that we can choose one for computation according to convenience and ease of calculation. We should do the calculations explicitly so that we can avoid making errors.