7.4-bo'lim : qisman kasrlar
6. \( \displaystyle \int{{\frac{{4x - 11}}{{{x^3} - 9{x^2}}}\,dx}}\) integralini baholang.
Yechim
Muammoni boshlash uchun bizga integralning qisman kasr parchalanishi shakli kerak. Biroq, buni olish uchun biz maxrajni hisobga olishimiz kerak.
\[\int{{\frac{{4x - 11}}{{{x^3} - 9{x^2}}}\,dx}} = \int{{\frac{{4x - 11}}{{{x^2}\left( {x - 9} \right)}}\,dx}}\]U holda integrand uchun qisman kasrning parchalanish shakli quyidagicha:
\[\frac{{4x - 11}}{{{x^2}\left( {x - 9} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 9}}\]2-qadam
Numeratorlarni teng belgilash,
\[4x - 11 = Ax\left( {x - 9} \right) + B\left( {x - 9} \right) + C\,{x^2}\]3-qadam
Biz ikkita koeffitsientni osongina olish uchun eslatmalarda muhokama qilingan "hiyla" dan foydalanishimiz mumkin va keyin uchinchisini olish uchun \(x\) ning boshqa qiymatini tanlashimiz mumkin, shuning uchun buni qilaylik. Mana o'sha ish.
\[\begin{align*}x = & \, 0 : & - 11 = & - 9B\\ & & & \\x = & \,9 : & 25 = &\, 81C\\ & & & \\x = & \,1 : & - 7 = & - 8A - 8B + C = - 8A - \frac{{767}}{{81}}\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned} & {A = - \frac{{25}}{{81}}}\\ & {B = \frac{{11}}{9}}\\ & {C = \frac{{25}}{{81}}}\end{aligned}\]Integratsiyaning qisman kasr shakli shunday bo'ladi:
\[\frac{{4x - 11}}{{{x^2}\left( {x - 9} \right)}} = \frac{{ - \frac{{25}}{{81}}}}{x} + \frac{{\frac{{11}}{9}}}{{{x^2}}} + \frac{{\frac{{25}}{{81}}}}{{x - 9}}\]4-qadam
Endi biz integralni bajarishimiz mumkin.
\[\int{{\frac{{4x - 11}}{{{x^2}\left( {x - 9} \right)}}\,dx}} = \int{{\frac{{ - \frac{{25}}{{81}}}}{x} + \frac{{\frac{{11}}{9}}}{{{x^2}}} + \frac{{\frac{{25}}{{81}}}}{{x - 9}}\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{25}}{{81}}\ln \left| x \right| - \frac{{\frac{{11}}{9}}}{x} + \frac{{25}}{{81}}\ln \left| {x - 9} \right| + c}}\]