Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 10.15 : Power Series and Functions
5. Give a power series representation for the integral of the following function.
\[h\left( x \right) = \frac{{{x^4}}}{{9 + {x^2}}}\]Show All Steps Hide All Steps
Hint : Integrating this function seems like (potentially) a lot of work, not to mention determining a power series representation of the result. It’s a good think that we know how to integrate power series.
First let’s notice that we can quickly find a power series representation for this function. Here is that work.
\[h\left( x \right) = \frac{{{x^4}}}{9}\frac{1}{{1 - \left( { - \frac{1}{9}{x^2}} \right)}} = \frac{{{x^4}}}{9}\sum\limits_{n = 0}^\infty {{{\left( { - \frac{1}{9}{x^2}} \right)}^n}} = \sum\limits_{n = 0}^\infty {\frac{1}{9}{x^4}{{\left( { - 1} \right)}^n}{{\left( {\frac{1}{9}} \right)}^n}\,{x^{2n}}} = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\left( {\frac{1}{9}} \right)}^{n + 1}}\,{x^{2n + 4}}} \] Show Step 2Now, we know how to integrate power series and we know that the integral of the power series representation of a function is the power series representation of the integral of the function.
Therefore,
\[\int{{h\left( x \right)\,dx}} = \int{{\sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\left( {\frac{1}{9}} \right)}^{n + 1}}\,{x^{2n + 4}}} \,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{c + \sum\limits_{n = 0}^\infty {\frac{1}{{2n + 5}}{{\left( { - 1} \right)}^n}{{\left( {\frac{1}{9}} \right)}^{n + 1}}\,{x^{2n + 5}}} }}\]Remember that to integrate a power series all we need to do is integrate the term of the power series and we can’t forget to add on the “+c” since we’re doing an indefinite integral.