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

\[u = t - 7\hspace{0.5in} \Rightarrow \hspace{0.5in}du = dt\hspace{0.25in}\& \hspace{0.25in}t = u + 7\]

O'zgartirishni amalga oshirish beradi,

\[\int{{\frac{{2t + 9}}{{{{\left( {{t^2} - 14t + 46} \right)}^{\frac{5}{2}}}}}\,dt}} = \int{{\frac{{2\left( {u + 7} \right) + 9}}{{{{\left( {{u^2} - 3} \right)}^{\frac{5}{2}}}}}\,du}} = \int{{\frac{{2u + 23}}{{{{\left( {{u^2} - 3} \right)}^{\frac{5}{2}}}}}\,du}}\]

Keyinchalik, biz integralni quyidagicha ajratishimiz kerak,

\[\int{{\frac{{2t + 9}}{{{{\left( {{t^2} - 14t + 46} \right)}^{\frac{5}{2}}}}}\,dt}} = \int{{\frac{{2u}}{{{{\left( {{u^2} - 3} \right)}^{\frac{5}{2}}}}}\,du}} + \int{{\frac{{23}}{{{{\left( {{u^2} - 3} \right)}^{\frac{5}{2}}}}}\,du}}\]

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

\[\begin{align*}\int{{\frac{{2t + 9}}{{{{\left( {{t^2} - 14t + 46} \right)}^{\frac{5}{2}}}}}\,dt}} & = \int{{{v^{ - \,\,\frac{5}{2}}}\,dv}} + \int{{\frac{{23}}{{{{\left( {3{{\sec }^2}\theta - 3} \right)}^{\frac{5}{2}}}}}\left( {\sqrt 3 \sec \theta \tan \theta } \right)\,d\theta }}\\ & = \int{{{v^{ - \,\,\frac{5}{2}}}\,dv}} + \int{{\frac{{23\sqrt 3 \sec \theta \tan \theta }}{{{{\left( {3{{\tan }^2}\theta } \right)}^{\frac{5}{2}}}}}\,d\theta }}\\ & = \int{{{v^{ - \,\,\frac{5}{2}}}\,dv}} + \int{{\frac{{23\sec \theta }}{{9{{\tan }^4}\theta }}\,d\theta }}\\ & = \int{{{v^{ - \,\,\frac{5}{2}}}\,dv}} + \frac{{23}}{9}\int{{\frac{{{{\cos }^3}\theta }}{{{{\sin }^4}\theta }}\,d\theta }}\end{align*}\]

Endi, ikkinchi integral uchun, biz tez-tez bajarishimiz kerak bo'lgan manipulyatsiyalarni unutmang, shuning uchun biz bunday integrallarni bajarishimiz mumkin. Agar sizga ushbu turdagi integrallar bo'yicha biroz mashq kerak bo'lsa, ushbu bobning ikkinchi bo'limi uchun amaliyot masalalariga qayting va ulardan ba'zilari bilan ishlang.

Bu muammo uchun integratsiya jarayonining qolgan qismi.

\[\begin{align*}\int{{\frac{{2t + 9}}{{{{\left( {{t^2} - 14t + 46} \right)}^{\frac{5}{2}}}}}\,dt}} & = \int{{{v^{ - \,\,\frac{5}{2}}}\,dv}} + \frac{{23}}{9}\int{{\frac{{1 - {{\sin }^2}\theta }}{{{{\sin }^4}\theta }}\cos \theta \,d\theta }}\hspace{0.25in}\,w = \sin \theta \\ & = \int{{{v^{ - \,\,\frac{5}{2}}}\,dv}} + \frac{{23}}{9}\int{{{w^{ - 4}} - {w^{ - 2}}\,dw}}\\ & = - \frac{2}{3}{v^{ - \,\,\frac{3}{2}}} + \frac{{23}}{9}\left[ { - \frac{1}{3}{{\left( {\sin \theta } \right)}^{ - 3}} + {{\left( {\sin \theta } \right)}^{ - 1}}} \right] + c\end{align*}\]

Javobni \(t\)-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, \[\sec \theta = \frac{u}{{\sqrt 3 }}\,\,\,\,\,\left( { = \frac{{{\mbox{hyp}}}}{{{\mbox{adj}}}}} \right)\] To'g'ri uchburchakdan biz olamiz, \[\sin \theta = \frac{{\sqrt {{u^2} - 3} }}{u}\]

Buni yuqoridagi integralga ulash natijasida,

\[\int{{\frac{{2t + 9}}{{{{\left( {{t^2} - 14t + 46} \right)}^{\frac{5}{2}}}}}\,dt}} = - \frac{2}{{3{{\left( {{u^2} - 3} \right)}^{\,\frac{3}{2}}}}} - \frac{{23{u^3}}}{{27{{\left( {{u^2} - 3} \right)}^{\frac{3}{2}}}}} + \frac{{23u}}{{9\sqrt {{u^2} - 3} }} + c\]

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.

\[\begin{align*}\int{{\frac{{2t + 9}}{{{{\left( {{t^2} - 14t + 46} \right)}^{\frac{5}{2}}}}}\,dt}} & = - \frac{2}{{3{{\left( {{{\left( {t - 7} \right)}^2} - 3} \right)}^{\frac{3}{2}}}}} - \frac{{23{{\left( {t - 7} \right)}^3}}}{{27{{\left( {{{\left( {t - 7} \right)}^2} - 3} \right)}^{\frac{3}{2}}}}} + \frac{{23\left( {t - 7} \right)}}{{9\sqrt {{{\left( {t - 7} \right)}^2} - 3} }} + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{23\left( {t - 7} \right)}}{{9\sqrt {{{\left( {t - 7} \right)}^2} - 3} }} - \frac{{18 + 23{{\left( {t - 7} \right)}^3}}}{{27{{\left( {{{\left( {t - 7} \right)}^2} - 3} \right)}^{\frac{3}{2}}}}} + c}}\end{align*}\]

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.