Chapter 7.3, Problem 7E

To determine

To find the indefinite integral.

Answer to Problem 7E

  $\text{The indefinite integral is, }{\displaystyle \int 3x^2{e}^{2x}dx}=\frac{3}{2}{x}^2{e}^{2x}-\frac{3}{2}x{e}^{2x}+\frac{3}{4}{e}^{2x}+C.$

Explanation of Solution

Given information: $\text{The equation is},{\displaystyle \int 3x^2{e}^{2x}dx}.$

Concept used: Tabular integration.

  $\begin{array}{l}\text{We have seen that integrals of the form }{\displaystyle \int f\left(x\right)}g\left(x\right)dx\text{ in which }f\text{ can be differentiated }\\ \text{repeatedly to become zero and }g\text{ can be integrated repeatedly without difficulty, are natural }\\ \text{candidates for integration by parts}\text{. However, if many repetitions are required, the calculation }\\ \text{can be cumbersome}\text{.}\end{array}$

Calculation : Using Tabular intergration, indefinite integral is obtained as,

  $\text{With }f\left(x\right)=3x^2\text{ and }g\left(x\right)=e^{2x},\text{ we list:}$

  

  $\begin{array}{l}\text{We combine the products of the functions connected by the arrows according to the }\\ \text{operation signs above the arrows to obtain,}\\ {\displaystyle \int 3x^2{e}^{2x}dx}=3x^2\left(\frac{1}{2}{e}^{2x}\right)-6x\left(\frac{1}{4}{e}^{2x}\right)+6\left(\frac{1}{8}{e}^{2x}\right)+C\\ =\frac{3}{2}{x}^2{e}^{2x}-\frac{3}{2}x{e}^{2x}+\frac{3}{4}{e}^{2x}+C\\ \text{Hence, the indefinite integral is, }{\displaystyle \int 3x^2{e}^{2x}dx}=\frac{3}{2}{x}^2{e}^{2x}-\frac{3}{2}x{e}^{2x}+\frac{3}{4}{e}^{2x}+C.\end{array}$