Chapter 6.3, Problem 45E

To determine

To find: The value of the integral.

Answer to Problem 45E

The answer is: $\frac{1}{2}{e}^{x^2}\left(x^6-3x^4+6x^2-6\right)+C$

Explanation of Solution

Given information:

The integral is ${{\displaystyle \int }}^{\text{}}{x}^7{e}^{x^2}dx$ .

Calculation:

  $\begin{array}{c}{{\displaystyle \int }}^{\text{}}{x}^7{e}^{x^2}dx={{\displaystyle \int }}^{\text{}}{\left(x^2\right)}^3{e}^{x^2}xdx\\ ={{\displaystyle \int }}^{\text{}}{u}^3{e}^u\left(\frac{1}{2}du\right)\\ =\frac{1}{2}{{\displaystyle \int }}^{\text{}}{u}^3{e}^{u}du\end{array}$

Since the derivative of $u^3$ needs to be taken multiple times to get to a constant, use tabular integration to evaluate ${{\displaystyle \int }}^{\text{}}{u}^3{e}^{u}du\text{.}$

Let ,

  $\begin{array}{c}f(x)=u^3\\ g(x)=e^u\end{array}$

Then,

  

Utilize the operation symbol located above each arrow to combine the results of the functions connected by arrows.

  $\begin{array}{c}{{\displaystyle \int }}^{\text{}}{u}^3{e}^{u}du=u^3{e}^u-3u^2{e}^u+6u{e}^u-6u^u+C\\ =e^u\left(u^3-3u^2+6u-6\right)+C\end{array}$

Insert the expression discovered for the integral before insert $u=x^2$ .

And then simplify,

  $\begin{array}{c}{{\displaystyle \int }}^{\text{}}{x}^7{e}^{x^2}dx=\frac{1}{2}{{\displaystyle \int }}^{\text{}}{u}^3{e}^{u}du\\ =\frac{1}{2}{e}^u\left(u^3-3u^2+6u-6\right)+C\\ =\frac{1}{2}{e}^{x^2}\left[{\left(x^2\right)}^3-3{\left(x^2\right)}^2+6x^2-6\right]+C\\ =\frac{1}{2}{e}^{x^2}\left(x^6-3x^4+6x^2-6\right)+C\end{array}$

Therefore, the required value of the given integral is $\frac{1}{2}{e}^{x^2}\left(x^6-3x^4+6x^2-6\right)+C$ .