Bundan ko'rishimiz mumkinki, quyidagi almashtirish biz uchun ishlashi kerak.

\[u = z + 1\hspace{0.5in} \Rightarrow \hspace{0.5in}du = dz\hspace{0.25in}\& \hspace{0.25in}z = u - 1\]

O'zgartirishni amalga oshirish beradi,

\[\int{{\frac{{3z}}{{{{\left( {1 - 4z - 2{z^2}} \right)}^2}}}\,dz}} = \int{{\frac{{3\left( {u - 1} \right)}}{{{{\left( {3 - 2{u^2}} \right)}^2}}}\,du}} = \int{{\frac{{3u - 3}}{{{{\left( {3 - 2{u^2}} \right)}^2}}}\,du}}\]

Keyinchalik, biz integralni quyidagicha ajratishimiz kerak,

\[\int{{\frac{{3z}}{{{{\left( {1 - 4z - 2{z^2}} \right)}^2}}}\,dz}} = \int{{\frac{{3u}}{{{{\left( {3 - 2{u^2}} \right)}^2}}}\,du}} - \int{{\frac{3}{{{{\left( {3 - 2{u^2}} \right)}^2}}}\,du}}\]

Birinchi integral \(v = 3 - 2{u^2}\) almashtirish bilan bajarilishi mumkin, ikkinchi integral esa \(u = \frac{{\sqrt 3 }}{{\sqrt 2 }}\sin \theta \) trig o'rnini bosishni talab qiladi. Mana almashtirish ishi.

\[\begin{align*}\int{{\frac{{3z}}{{{{\left( {1 - 4z - 2{z^2}} \right)}^2}}}\,dz}} & = - \frac{3}{4}\int{{{v^{ - 2}}\,dv}} - \int{{\frac{3}{{{{\left( {3 - 3{{\sin }^2}\theta } \right)}^2}}}\left( {\frac{{\sqrt 3 }}{{\sqrt 2 }}\cos \theta } \right)\,d\theta }}\\ & = - \frac{3}{4}\int{{{v^{ - 2}}\,dv}} - \int{{\frac{3}{{{{\left( {3{{\cos }^2}\theta } \right)}^2}}}\left( {\frac{{\sqrt 3 }}{{\sqrt 2 }}\cos \theta } \right)\,d\theta }}\\ & = - \frac{3}{4}\int{{{v^{ - 2}}\,dv}} - \frac{1}{{\sqrt 6 }}\int{{{{\sec }^3}\theta \,d\theta }}\end{align*}\]

Ushbu muammo uchun ikkinchi integral ushbu bobning ikkinchi qismi uchun eslatmalarda bajarilgan integralga to'g'ri keladi va shuning uchun biz ushbu integralni bajarish uchun ushbu bo'limda olingan formuladan foydalanamiz.

Bu muammo uchun integratsiya jarayonining qolgan qismi.

\[\int{{\frac{{3z}}{{{{\left( {1 - 4z - 2{z^2}} \right)}^2}}}\,dz}} = \frac{3}{4}{v^{ - 1}} - \frac{1}{{2\sqrt 6 }}\left[ {\sec \theta \tan \theta + \ln \left| {\sec \theta + \tan \theta } \right|} \right] + c\]

Javobni \(z\)-ga qaytarish uchun endi biz biroz orqaga almashtirishimiz kerak. Ikkinchi integratsiya natijasi bilan boshlaylik. \(\theta \)-ning orqa qismini \(u\)-ga aylantirish uchun tezkor uchburchak kerak bo'ladi.

Bizda mavjud almashtirishdan, \[\sin \theta = \frac{{\sqrt 2 \,u}}{{\sqrt 3 }}\,\,\,\,\,\left( { = \frac{{{\mbox{opp}}}}{{{\mbox{hyp}}}}} \right)\] To'g'ri uchburchakdan biz olamiz, \[\tan \theta = \frac{{\sqrt 2 \,u}}{{\sqrt {3 - 2{u^2}} }}\,\,\,\,\,\,\,\& \,\,\,\,\,\,\,\sec \theta = \frac{{\sqrt 3 }}{{\sqrt {3 - 2{u^2}} }}\]

Buni yuqoridagi integralga ulash natijasida,

\[\int{{\frac{{3z}}{{{{\left( {1 - 4z - 2{z^2}} \right)}^2}}}\,dz}} = \frac{3}{{4\left( {3 - 2{u^2}} \right)}} - \frac{1}{{2\sqrt 6 }}\left[ {\frac{{\sqrt 6 \,u}}{{3 - 2{u^2}}} + \ln \left| {\frac{{\sqrt 3 + \sqrt 2 \,u}}{{\sqrt {3 - 2{u^2}} }}} \right|} \right] + c\]

E'tibor bering, biz birinchi muddatda ham \(v\) ni almashtirdik va birinchi atamani biroz qayta yozdik. Va nihoyat, biz qilishimiz kerak bo'lgan yagona narsa \(u\) o'rnini bosishdir.

\[\int{{\frac{{3z}}{{{{\left( {1 - 4z - 2{z^2}} \right)}^2}}}\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{3}{{4\left( {3 - 2{{\left( {z + 1} \right)}^2}} \right)}} - \frac{{z + 1}}{{6 - 4{{\left( {z + 1} \right)}^2}}} - \frac{1}{{2\sqrt 6 }}\ln \left| {\frac{{\sqrt 3 + \sqrt 2 \,\left( {z + 1} \right)}}{{\sqrt {3 - 2{{\left( {z + 1} \right)}^2}} }}} \right| + c}}\]

Biz ushbu yechimni ushbu turdagi muammolar haqida yakuniy eslatma bilan qoldiramiz. Ular ko'pincha juda uzun, tartibsiz va xatolar uchun keng imkoniyatlar mavjud, shuning uchun ular bilan ehtiyot bo'ling va ular bilan ishlashda juda ko'p shoshilmang.