Endi biz almashtirishga tayyorlanishimiz kerak. Boshqacha qilib aytganda, biz \(t\) ni hal qilishimiz va \(dt\) ni olishimiz kerak.

\[t = \frac{1}{2}{u^2} + 2\hspace{0.5in} \Rightarrow \hspace{0.5in}dt = u\,du\]

O'zgartirishni amalga oshirish beradi,

\[\int{{\frac{{t - 2}}{{t - 3\sqrt {2t - 4} + 2}}\,dt}} = \int{{\frac{{\frac{1}{2}{u^2} + 2 - 2}}{{\frac{1}{2}{u^2} + 2 - 3u + 2}}\,\left( u \right)du}} = \int{{\frac{{{u^3}}}{{{u^2} - 6u + 8}}\,du}}\]

Bu integral baholash uchun qisman kasrlarni talab qiladi.

Biroq, biz birinchi navbatda integratsiya bo'yicha uzun bo'linishimiz kerak, chunki pay (3) darajasi maxraj (2) darajasidan yuqori. Bu beradi,

\[\frac{{{u^3}}}{{{u^2} - 6u + 8}} = u + 6 + \frac{{28u - 48}}{{\left( {u - 2} \right)\left( {u - 4} \right)}}\]

Uchinchi hadda qisman kasrning parchalanish shakli:

\[\frac{{28u - 48}}{{\left( {u - 2} \right)\left( {u - 4} \right)}} = \frac{A}{{u - 2}} + \frac{B}{{u - 4}}\]

Koeffitsientlarni teng belgilash,

\[28u - 48 = A\left( {u - 4} \right) + B\left( {u - 2} \right)\]

Koeffitsientlarni olish uchun "hiyla" dan foydalanish,

\[\begin{align*}{u = 4:} & \hspace{0.25in}& 64 = & \, 2B\\{u = 2:} & \hspace{0.25in} & 8 = & - 2A\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in} \begin{aligned} & {A = - 4}\\ & {B = 32}\end{aligned}\]

U holda integral,

\[\int{{\frac{{{u^3}}}{{{u^2} - 6u + 8}}\,du}} = \int{{u + 6 - \frac{4}{{u - 2}} + \frac{{32}}{{u - 4}}\,du}} = \frac{1}{2}{u^2} + 6u - 4\ln \left| {u - 2} \right| + 32\ln \left| {u - 4} \right| + c\]

Oxirgi qadam, yakuniy javobni olish uchun barcha orqadagi almashtirishlarni bajarishdir.

\[\int{{\frac{{{u^3}}}{{{u^2} - 6u + 8}}\,du}} = \require{bbox} \bbox[2pt,border:1px solid black]{{t - 2 + 6\sqrt {2t - 4} - 4\ln \left| {\sqrt {2t - 4} - 2} \right| + 32\ln \left| {\sqrt {2t - 4} - 4} \right| + c}}\]