Numeratorlarni teng belgilash,

\[\begin{align*}8 + t + 6{t^2} - 12{t^3} & = \left( {At + B} \right)\left( {{t^2} + 7} \right) + \left( {Ct + D} \right)\left( {3{t^2} + 4} \right)\\ & = \left( {A + 3C} \right){t^3} + \left( {B + 3D} \right){t^2} + \left( {7A + 4C} \right)t + 7B + 4D\end{align*}\]

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.

Endi koeffitsientlarni teng qo'yish quyidagi tizimni beradi.

\[\begin{align*}{{t^3}:} & \hspace{0.25in} & A + 3C = & - 12\\{{t^2}:} & \hspace{0.25in} & B + 3D = & \, 6\\{{t^1}:} & \hspace{0.25in} & 7A + 4C = &\, 1\\{{t^0}:} & \hspace{0.25in} & 7B + 4D = &\, 8\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned} & {A = 3}\\ & {B = 0}\\ & {C = - 5}\\ & {D = 2}\end{aligned}\]

Integratsiyaning qisman kasr shakli shunday bo'ladi:

\[\frac{{8 + t + 6{t^2} - 12{t^3}}}{{\left( {3{t^2} + 4} \right)\left( {{t^2} + 7} \right)}} = \frac{{3t}}{{3{t^2} + 4}} + \frac{{ - 5t + 2}}{{{t^2} + 7}}\]

Endi biz integralni bajarishimiz mumkin.

\[\begin{align*}\int{{\frac{{8 + t + 6{t^2} - 12{t^3}}}{{\left( {3{t^2} + 4} \right)\left( {{t^2} + 7} \right)}}\,dt}} & = \int{{\frac{{3t}}{{3{t^2} + 4}} + \frac{{ - 5t + 2}}{{{t^2} + 7}}\,dt}}\\ & = \int{{\frac{{3t}}{{3{t^2} + 4}} - \frac{{5t}}{{{t^2} + 7}} + \frac{2}{{{t^2} + 7}}\,dt}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{2}\ln \left| {3{t^2} + 4} \right| - \frac{5}{2}\ln \left| {{t^2} + 7} \right| + \frac{2}{{\sqrt 7 }}{{\tan }^{ - 1}}\left( {\frac{t}{{\sqrt 7 }}} \right) + c}}\end{align*}\]

E'tibor bering, birinchi va ikkinchi integratsiya uchun mos ravishda \(u = 3{t^2} + 4\) va \(u = {t^2} + 7\) almashtirishlari kerak bo'lsa, uchinchisi esa eslatmalarda keltirilgan formulaga muhtoj edi.