7.4-bo'lim : qisman kasrlar
10. \( \displaystyle \int{{\frac{{2 + {w^4}}}{{{w^3} + 9w}}\,dw}}\) integralini baholang.
Yechim
Esda tutingki, agar ayiruvchining darajasi maxraj darajasidan kichik bo'lsa, biz ratsional ifodada qisman kasrlarni bajarishimiz mumkin. Bunda hisobning darajasi 4 ga, maxrajning darajasi esa 3 ga teng.
Shunday qilib, birinchi qadam - uzun bo'linish (uzoq bo'linish uchun Algebra ko'nikmalarimizni tekshirishni sizga qoldiramiz) olish uchun,
\[\frac{{2 + {w^4}}}{{{w^3} + 9w}} = w + \frac{{2 - 9{w^2}}}{{w\left( {{w^2} + 9} \right)}}\]2-qadam
Endi ikkinchi hadda qisman kasrlarni bajarishimiz mumkin. Bu erda qisman kasr parchalanish shakli.
\[\frac{{2 - 9{w^2}}}{{w\left( {{w^2} + 9} \right)}} = \frac{A}{w} + \frac{{Bw + C}}{{{w^2} + 9}}\]Numeratorlarni teng belgilash,
\[2 - 9{w^2} = A\left( {{w^2} + 9} \right) + w\left( {Bw + C} \right) = \left( {A + B} \right){w^2} + Cw + 9A\]Bunday holda, eslatmalarda muhokama qilingan "hiyla" biz uchun unchalik yaxshi ishlamaydi va shuning uchun biz hamma narsani ko'paytirishga va yuqorida ko'rsatilgandek atamalarni yig'ishga murojaat qilishimiz kerak.
3-qadam
Endi koeffitsientlarni teng qo'yish quyidagi tizimni beradi.
\[\begin{align*}{{w^2}:} & \hspace{0.25in} & A + B = & - 9\\ & & & \\{{w^1}:} & \hspace{0.25in} & C = & \,0\\ & & & \\{{w^0}:} & \hspace{0.25in} & 9A = &\, 2\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned} & {A = \frac{2}{9}}\\ & {B = - \frac{{83}}{9}}\\ & {C = 0}\end{aligned}\]Ikkinchi hadning qisman kasr shakli u holda,
\[\frac{{2 - 9{w^2}}}{{w\left( {{w^2} + 9} \right)}} = \frac{{\frac{2}{9}}}{w} - \frac{{\frac{{83}}{9}w}}{{{w^2} + 9}}\]4-qadam
Endi biz integralni bajarishimiz mumkin.
\[\int{{\frac{{2 + {w^4}}}{{{w^3} + 9w}}\,dw}} = \int{{w + \frac{{\frac{2}{9}}}{w} - \frac{{\frac{{83}}{9}w}}{{{w^2} + 9}}\,dw}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{2}{w^2} + \frac{2}{9}\ln \left| w \right| - \frac{{83}}{{18}}\ln \left| {{w^2} + 9} \right| + c}}\]