7.4-bo'lim : qisman kasrlar
7. \( \displaystyle \int{{\frac{{{z^2} + 2z + 3}}{{\left( {z - 6} \right)\left( {{z^2} + 4} \right)}}\,dz}}\) integralini baholang.
Yechim
Bunday holda, maxraj allaqachon faktorlarga ajratilgan va shuning uchun biz to'g'ridan-to'g'ri integratsiya uchun qisman kasr parchalanish shakliga o'tishimiz mumkin.
\[\frac{{{z^2} + 2z + 3}}{{\left( {z - 6} \right)\left( {{z^2} + 4} \right)}} = \frac{A}{{z - 6}} + \frac{{Bz + C}}{{{z^2} + 4}}\]2-qadam
Numeratorlarni teng belgilash,
\[{z^2} + 2z + 3 = A\left( {{z^2} + 4} \right) + \left( {Bz + C} \right)\left( {z - 6} \right) = \left( {A + B} \right){z^2} + \left( { - 6B + C} \right)z + 4A - 6C\]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.
\[\begin{align*}{{z^2}:}& \hspace{0.25in} & A + B = & \, 1\\ & \\ {{z^1}:}& \hspace{0.25in}& - 6B + C = & \, 2\\ & \\{{z^0}:}& \hspace{0.25in} & 4A - 6C = & \, 3\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\begin{aligned}&{A = \frac{{51}}{{40}}}\\&{B = - \frac{{11}}{{40}}}\\&{C = \frac{7}{{20}}}\end{aligned}\]Integratsiyaning qisman kasr shakli shunday bo'ladi:
\[\frac{{{z^2} + 2z + 3}}{{\left( {z - 6} \right)\left( {{z^2} + 4} \right)}} = \frac{{\frac{{51}}{{40}}}}{{z - 6}} + \frac{{ - \frac{{11}}{{40}}z + \frac{7}{{20}}}}{{{z^2} + 4}}\]4-qadam
Endi biz integralni bajarishimiz mumkin.
\[\begin{align*}\int{{\frac{{{z^2} + 2z + 3}}{{\left( {z - 6} \right)\left( {{z^2} + 4} \right)}}\,dz}} & = \int{{\frac{{\frac{{51}}{{40}}}}{{z - 6}} + \frac{{ - \frac{{11}}{{40}}z + \frac{7}{{20}}}}{{{z^2} + 4}}\,dz}}\\ & = \int{{\frac{{\frac{{51}}{{40}}}}{{z - 6}} + \frac{{ - \frac{{11}}{{40}}z}}{{{z^2} + 4}} + \frac{{\frac{7}{{20}}}}{{{z^2} + 4}}\,dz}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{51}}{{40}}\ln \left| {z - 6} \right| - \frac{{11}}{{80}}\ln \left| {{z^2} + 4} \right| + \frac{7}{{40}}{{\tan }^{ - 1}}\left( {\frac{z}{2}} \right) + c}}\end{align*}\]E'tibor bering, ikkinchi integratsiya \(u = {z^2} + 4\) o'rnini, uchinchisiga esa eslatmalarda keltirilgan formula kerak edi.