Avval shuni ta'kidlaymizki, biz ushbu funktsiya uchun quvvat seriyasining tasvirini tezda topishimiz mumkin. Mana o'sha ish.

\[h\left( x \right) = \frac{{{x^4}}}{9}\frac{1}{{1 - \left( { - \frac{1}{9}{x^2}} \right)}} = \frac{{{x^4}}}{9}\sum\limits_{n = 0}^\infty {{{\left( { - \frac{1}{9}{x^2}} \right)}^n}} = \sum\limits_{n = 0}^\infty {\frac{1}{9}{x^4}{{\left( { - 1} \right)}^n}{{\left( {\frac{1}{9}} \right)}^n}\,{x^{2n}}} = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\left( {\frac{1}{9}} \right)}^{n + 1}}\,{x^{2n + 4}}} \]

2-qadam

Endi biz darajali qatorlarni qanday integrallashni bilamiz va biz bilamizki, funktsiyaning darajali qator ko'rinishining integrali funksiya integralining darajali qator ko'rinishidir.

Shuning uchun,

\[\int{{h\left( x \right)\,dx}} = \int{{\sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\left( {\frac{1}{9}} \right)}^{n + 1}}\,{x^{2n + 4}}} \,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{c + \sum\limits_{n = 0}^\infty {\frac{1}{{2n + 5}}{{\left( { - 1} \right)}^n}{{\left( {\frac{1}{9}} \right)}^{n + 1}}\,{x^{2n + 5}}} }}\]

Esda tutingki, kuch seriyasini birlashtirish uchun biz qilishimiz kerak bo'lgan narsa - kuch seriyasining terminini birlashtirish va biz "+c" ni qo'shishni unuta olmaymiz, chunki biz noaniq integral qilamiz.