7.4-bo'lim : qisman kasrlar
9. \( \displaystyle \int{{\frac{{6{x^2} - 3x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}}\,dx}}\) integralini baholang.
Yechim
Esda tutingki, agar ayiruvchining darajasi maxraj darajasidan kichik bo'lsa, biz ratsional ifodada qisman kasrlarni bajarishimiz mumkin. Bu holda pay ham, maxraj ham 2 daraja bo'ladi. Agar maxrajni ko'paytirsak, buni osongina ko'rish mumkin.
\[\frac{{6{x^2} - 3x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}} = \frac{{6{x^2} - 3x}}{{{x^2} + 2x - 8}}\]Shunday qilib, birinchi qadam - uzun bo'linish (uzoq bo'linish uchun Algebra ko'nikmalarimizni tekshirishni sizga qoldiramiz) olish uchun,
\[\frac{{6{x^2} - 3x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}} = 6 + \frac{{48 - 15x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}}\]2-qadam
Endi ikkinchi hadda qisman kasrlarni bajarishimiz mumkin. Bu erda qisman kasr parchalanish shakli.
\[\frac{{48 - 15x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}} = \frac{A}{{x - 2}} + \frac{B}{{x + 4}}\]Numeratorlarni teng belgilash,
\[48 - 15x = A\left( {x + 4} \right) + B\left( {x - 2} \right)\]3-qadam
Bu erda "hiyla" oson ishlaydi, shuning uchun bu ish.
\[\begin{align*}{x = - 4:} & \hspace{0.25in} & 108 = & - 6B\\{x = 2:} & \hspace{0.25in} & 18 = & \, 6A\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned} & {A = 3}\\ & {B = - 18}\end{aligned}\]Ikkinchi hadning qisman kasr shakli u holda,
\[\frac{{48 - 15x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}} = \frac{3}{{x - 2}} - \frac{{18}}{{x + 4}}\]4-qadam
Endi biz integralni bajarishimiz mumkin.
\[\int{{\frac{{6{x^2} - 3x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}}\,dx}} = \int{{6 + \frac{3}{{x - 2}} - \frac{{18}}{{x + 4}}\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{6x + 3\ln \left| {x - 2} \right| - 18\ln \left| {x + 4} \right| + c}}\]