7.4-bo'lim: Qisman kasrlar
Ushbu bo'limda biz ko'phadlarning ratsional ifodalarining integrallarini ko'rib chiqamiz va yana bir bor ushbu bo'limni biz allaqachon bajara oladigan integral bilan boshlaymiz, shuning uchun uni ushbu bo'limda bajariladigan integrallarga qarama-qarshi qo'yishimiz mumkin.
\[\begin{align*}\int{{\frac{{2x - 1}}{{{x^2} - x - 6}}\,dx}} & = \int{{\frac{1}{u}\,du}}\hspace{0.25in}{\mbox{using }}u = {x^2} - x - 6\,\,{\mbox{ and }}du = \left( {2x - 1} \right)dx\\ & = \ln \left| {{x^2} - x - 6} \right| + c\end{align*}\]Demak, agar pay maxrajning hosilasi (yoki maxraj hosilasining doimiy karrali) bo'lsa, bunday integralni bajarish juda oddiy. Biroq, ko'pincha hisoblagich maxrajning hosilasi emas (yoki doimiy ko'paytma). Masalan, quyidagi integralni ko'rib chiqing.
\[\int{{\frac{{3x + 11}}{{{x^2} - x - 6}}\,dx}}\]Bu holda pay aniq maxrajning hosilasi emas va maxrajning hosilasining doimiy karralisi ham emas. Shuning uchun biz yuqorida ishlatgan oddiy almashtirish ishlamaydi. Ammo, agar biz integralni quyidagicha ajratish mumkinligini sezsak,
\[\frac{{3x + 11}}{{{x^2} - x - 6}} = \frac{4}{{x - 3}} - \frac{1}{{x + 2}}\]u holda integral aslida juda oddiy.
\[\begin{align*}\int{{\frac{{3x + 11}}{{{x^2} - x - 6}}\,dx}} & = \int{{\frac{4}{{x - 3}}\, - \frac{1}{{x + 2}}dx}}\\ & = 4\ln \left| {x - 3} \right| - \ln \left| {x + 2} \right| + c\end{align*}\]Ratsional ifodani olish va uni oddiyroq ratsional iboralarga ajratish jarayoni, biz asl ratsional ifodani olish uchun qo‘shish yoki ayirish mumkin bo‘lgan jarayon deyiladi. qisman fraksiya parchalanishi. Ratsional ifodalarni o'z ichiga olgan ko'plab integrallarni bajarish mumkin, agar biz birinchi navbatda integralda qisman kasrlarni bajarsak.
Shunday qilib, keling, qisman kasrlarni tez ko'rib chiqaylik. Biz shaklda ratsional ifoda bilan boshlaymiz,
\[f\left( x \right) = \frac{{P\left( x \right)}}{{Q\left( x \right)}}\]bu erda \(P\left( x \right)\) ham, \(Q\left( x \right)\) ham polinomlar va \(P\left( x \right)\) darajasi \(Q\left( x \right)\) darajasidan kichikroq. Eslatib o'tamiz, ko'phadning darajasi ko'phaddagi eng katta ko'rsatkichdir. Qisman kasrlar faqat payning darajasi maxrajning darajasidan qat'iy kam bo'lgan taqdirdagina amalga oshirilishi mumkin. Buni eslash muhim.
Shunday qilib, biz qisman kasrlarni bajarish mumkinligini aniqlaganimizdan so'ng, maxrajni iloji boricha to'liq koeffitsientga olamiz. Keyin maxrajdagi har bir omil uchun biz qisman kasr parchalanishida oladigan atama(lar)ni aniqlash uchun quyidagi jadvaldan foydalanishimiz mumkin.
| Faktor maxraj |
Muddat qisman fraksiyalarning parchalanishi |
|---|---|
| \( \displaystyle ax + b\) | \( \displaystyle \frac{A}{{ax + b}}\) |
| \( \displaystyle {\left( {ax + b} \right)^k}\) | \( \displaystyle \frac{{{A_1}}}{{ax + b}} + \frac{{{A_2}}}{{{{\left( {ax + b} \right)}^2}}} + \cdots + \frac{{{A_k}}}{{{{\left( {ax + b} \right)}^k}}}\), \(k = 1,2,3, \ldots \) |
| \( \displaystyle a{x^2} + bx + c\) | \( \displaystyle \frac{{Ax + B}}{{a{x^2} + bx + c}}\) |
| \( \displaystyle {\left( {a{x^2} + bx + c} \right)^k}\) | \( \displaystyle \frac{{{A_1}x + {B_1}}}{{a{x^2} + bx + c}} + \frac{{{A_2}x + {B_2}}}{{{{\left( {a{x^2} + bx + c} \right)}^2}}} + \cdots + \frac{{{A_k}x + {B_k}}}{{{{\left( {a{x^2} + bx + c} \right)}^k}}}\), \(k = 1,2,3, \ldots \) |
E'tibor bering, birinchi va uchinchi holatlar, albatta, mos ravishda ikkinchi va to'rtinchi holatlarning alohida holatlaridir.
Har bir atama uchun koeffitsientlarni aniqlashning bir necha usullari mavjud va biz quyidagi misollarda ularning har birini ko'rib chiqamiz.
Yuqoridagi integralni bajarish bilan misollarni boshlaylik.
Yechim
Birinchi qadam maxrajni iloji boricha koeffitsientga ajratish va qisman kasr parchalanish shaklini olishdir. Buni qilish,
\[\frac{{3x + 11}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\, = \frac{A}{{x - 3}} + \frac{B}{{x + 2}}\]Keyingi qadam, aslida o'ng tomonni zaxiraga qo'shishdir.
\[\frac{{3x + 11}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\, = \frac{{A\left( {x + 2} \right) + B\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\]Endi biz \(A\) va \(B\) ni tanlashimiz kerak, shunda bu ikkalasining suratlari har bir \(x\) uchun teng bo'ladi. Buning uchun biz hisoblagichlarni tenglashtirishimiz kerak.
\[3x + 11 = A\left( {x + 2} \right) + B\left( {x - 3} \right)\]E'tibor bering, ko'pgina muammolarda biz parchalanishning umumiy shaklidan to'g'ridan-to'g'ri ushbu bosqichga o'tamiz va atamalarni zaxira nusxasini qo'shish bilan bezovtalanmaymiz. Shartlarni qo'shishning yagona nuqtasi hisoblagichni olishdir va biz buni qo'shish natijalarini yozmasdan ham olishimiz mumkin.
Shu nuqtada bizda davom etishning ikkita usulidan biri bor. Bir yo'l har doim ishlaydi, lekin ko'pincha ko'proq ishlaydi. Ikkinchisi, har doim ham ishlamasa-da, ishlaganda tez-tez ishlaydi. Bu holda ikkalasi ham ishlaydi va shuning uchun biz ushbu misol uchun tezroq usuldan foydalanamiz. Keyingi misolda boshqa usulni ko'rib chiqamiz.
Bu erda biz hisoblagichlar teng bo'lishi kerakligini payqashimiz kerak har qanday x biz foydalanishni tanlaymiz. Xususan, suratlar \(x = - 2\) va \(x = 3\) uchun teng bo'lishi kerak. Shunday qilib, keling, bularni ulang va nima olishimizni ko'rib chiqaylik.
\[\begin{align*}x & = - 2 : & \hspace{0.5in}5 & = A\left( 0 \right) + B\left( { - 5} \right) & \hspace{0.25in} & \Rightarrow & \hspace{0.25in}B & = - 1\\ x & = 3 \,\,\,\,: & \hspace{0.5in}20 & = A\left( 5 \right) + B\left( 0 \right) & \hspace{0.25in} & \Rightarrow & \hspace{0.25in}A & = 4\end{align*}\]Shunday qilib, \(x\)-ni sinchkovlik bilan tanlab, biz tezda chiqib ketish uchun noma'lum konstantalarni oldik. E'tibor bering, bu biz yuqorida bo'lishini da'vo qilgan qiymatlar.
Hozirgi vaqtda integraldan boshqa ko'p narsa qilish kerak emas.
\[\begin{align*}\int{{\frac{{3x + 11}}{{{x^2} - x - 6}}\,dx}} & = \int{{\frac{4}{{x - 3}}\, - \frac{1}{{x + 2}}dx}}\\ & = \int{{\frac{4}{{x - 3}}\,dx}} - \int{{\frac{1}{{x + 2}}dx}}\\ & = 4\ln \left| {x - 3} \right| - \ln \left| {x + 2} \right| + c\end{align*}\]Eslatib o'tamiz, bu integralni bajarish uchun avval uni ikkita integralga ajratamiz va keyin almashtirishlardan foydalanamiz,
\[u = x - 3\hspace{0.5in}v = x + 2\]yakuniy javobni olish uchun integrallar haqida.
Keyingi misolga o'tishdan oldin, bu erda bir nechta tezkor eslatmalar mavjud. Birinchidan, qisman kasr masalalaridagi ko'pgina integrallar yuqorida ko'rsatilgan integral turiga to'g'ri keladi. Ushbu integrallarni bajarishingiz mumkinligiga ishonch hosil qiling.
Bundan tashqari, bu kabi muammolarda tez-tez namoyon bo'ladigan yana bir integral mavjud, shuning uchun biz bu erda uning formulasini keltirishimiz mumkin, chunki biz allaqachon mavzudamiz.
\[\int{{\frac{1}{{{x^2} + {a^2}}}\,dx}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + c\]Buni ishlatishdan oldin bir yoki ikkita misol bo'ladi, shuning uchun bu haqda unutmang.
Endi yana bir nechta misollar bilan ishlaylik.
Yechim
Biz oldingi misolda bo'lgani kabi, bu yechimga unchalik ko'p ma'lumot bermaymiz. Birinchi narsa, maxrajni faktorlarga ajratish va qisman kasr parchalanish shaklini olishdir.
\[\frac{{{x^2} + 4}}{{x\left( {x + 2} \right)\left( {3x - 2} \right)}}\, = \frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{3x - 2}}\]Keyingi qadam hisoblagichlarni tenglashtirishdir. Agar siz o'sha tomon uchun hisoblagichni olish uchun haqiqatan ham o'ng tomonni qo'shishingiz kerak bo'lsa, unda buni qilishingiz kerak, ammo agar siz qo'shishni boshingizga qo'shsangiz, bu muammoni tezroq hal qiladi.
\[{x^2} + 4 = A\left( {x + 2} \right)\left( {3x - 2} \right) + Bx\left( {3x - 2} \right) + Cx\left( {x + 2} \right)\]Oldingi misolda bo'lgani kabi, biz \(x\) ning bir nechta qiymatlarini tanlashimiz va doimiylarni topishimiz mumkin, shuning uchun buni qilaylik.
\[\begin{align*}x & = 0 \,\,\,\,\, : & \hspace{0.5in}4 & = A\left( 2 \right)\left( { - 2} \right) & \hspace{0.5in} & \Rightarrow & \hspace{0.25in}A & = - 1\\ x & = - 2 : & \hspace{0.5in}8 & = B\left( { - 2} \right)\left( { - 8} \right) & \hspace{0.25in}&\Rightarrow & \hspace{0.25in}B & = \frac{1}{2}\\ x & = \frac{2}{3}\,\, : & \hspace{0.5in}\frac{{40}}{9} & = C\left( {\frac{2}{3}} \right)\left( {\frac{8}{3}} \right) & \hspace{0.25in} & \Rightarrow & \hspace{0.25in}C & = \frac{{40}}{{16}} = \frac{5}{2}\end{align*}\]E'tibor bering, birinchi misoldan farqli o'laroq, bu erda koeffitsientlarning aksariyati kasrlardir. Bu g'ayrioddiy emas, shuning uchun bu sodir bo'lganda hayajonlanmang.
Endi integralni bajaramiz.
\[\begin{align*}\int{{\frac{{{x^2} + 4}}{{3{x^3} + 4{x^2} - 4x}}\,dx}} & = \int{{ - \frac{1}{x} + \frac{{\frac{1}{2}}}{{x + 2}} + \frac{{\frac{5}{2}}}{{3x - 2}}\,dx}}\\ & = - \ln \left| x \right| + \frac{1}{2}\ln \left| {x + 2} \right| + \frac{5}{6}\ln \left| {3x - 2} \right| + c\end{align*}\]Shunga qaramay, yuqorida aytib o'tilganidek, tabiiy logarifmlarni yaratuvchi integrallar bu muammolarda juda keng tarqalgan, shuning uchun ularni bajarishingiz mumkinligiga ishonch hosil qiling. Bundan tashqari, siz oxirgi integralni to'g'ri bajardingizmi? \(\frac{5}{6}\) koeffitsienti to'g'ri. Termin uchun zarur bo'lgan almashtirishni to'g'ri bajarganingizga ishonch hosil qiling.
Yechim
Bu safar maxraj faktorlarga ajratilgan, shuning uchun to'g'ridan-to'g'ri qisman kasr parchalanishiga o'tamiz.
\[\frac{{{x^2} - 29x + 5}}{{{{\left( {x - 4} \right)}^2}\left( {{x^2} + 3} \right)}}\, = \frac{A}{{x - 4}} + \frac{B}{{{{\left( {x - 4} \right)}^2}}} + \frac{{Cx + D}}{{{x^2} + 3}}\]Numeratorlarni o'rnatish,
\[{x^2} - 29x + 5 = A\left( {x - 4} \right)\left( {{x^2} + 3} \right) + B\left( {{x^2} + 3} \right) + \left( {Cx + D} \right){\left( {x - 4} \right)^2}\]Bunday holda, biz faqat \(x\) qiymatlarini tanlay olmaymiz, bu bizga barcha konstantalarni beradi. Shuning uchun, biz buni ikkinchi (va ko'pincha uzoqroq) usulda ishlashimiz kerak bo'ladi. Birinchi qadam - o'ng tomonni ko'paytirish va barcha shunga o'xshash atamalarni bir joyga to'plash. Buni qilish,
\[{x^2} - 29x + 5 = \left( {A + C} \right){x^3} + \left( { - 4A + B - 8C + D} \right){x^2} + \left( {3A + 16C - 8D} \right)x - 12A + 3B + 16D\]Endi biz \(A\), \(B\), \(C\) va \(D\) ni tanlashimiz kerak, shunda bu ikkalasi teng bo'ladi. Boshqacha qilib aytganda, biz \(x\) ning o'xshash kuchlari koeffitsientlarini tenglashtirishimiz kerak. Bu yechish mumkin bo'lgan tenglamalar sistemasini beradi.
\[\left. \begin{aligned}{x^3} & :\hspace{0.25in} & A + C & = 0\\ {x^2} & :\hspace{0.25in} & - 4A + B - 8C + D & = 1\\ {x^1} & :\hspace{0.25in} & 3A + 16C - 8D & = - 29\\ {x^0} & :\hspace{0.25in} & - 12A + 3B + 16D & = 5\end{aligned} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}A = 1,\,B = - 5,\,C = - 1,\,D = 2\]E'tibor bering, biz konstantalarni ifodalash uchun \({x^0}\) dan foydalanganmiz. Shuni ham yodda tutingki, ushbu tizimlar ko'pincha juda katta bo'lishi mumkin va ularni hal qilishda adolatli ish hajmi mavjud. Ularni hal qilishning eng yaxshi usuli kompyuter yordamida hal qilish usullaridan foydalanishdir.
Endi integralni ko'rib chiqaylik.
\[\begin{align*}\int{{\frac{{{x^2} - 29x + 5}}{{{{\left( {x - 4} \right)}^2}\left( {{x^2} + 3} \right)}}\,dx}} & = \int{{\frac{1}{{x - 4}} - \frac{5}{{{{\left( {x - 4} \right)}^2}}} + \frac{{ - x + 2}}{{{x^2} + 3}}\,dx}}\\ & = \int{{\frac{1}{{x - 4}} - \frac{5}{{{{\left( {x - 4} \right)}^2}}} - \frac{x}{{{x^2} + 3}}\, + \frac{2}{{{x^2} + 3}}\,dx}}\\ & = \ln \left| {x - 4} \right| + \frac{5}{{x - 4}} - \frac{1}{2}\ln \left| {{x^2} + 3} \right| + \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 3 }}} \right) + c\end{align*}\]Uchinchi muddatga g'amxo'rlik qilish uchun biz uni ikkita alohida atamaga bo'lishimiz kerak edi. Buni bajarganimizdan so'ng, biz muammodagi barcha integrallarni bajarishimiz mumkin. Birinchi ikkitasi \(u = x - 4\) almashtirishdan, uchinchisi \(v = {x^2} + 3\) o'rnini egallaydi va to'rtinchi atama teskari tangenslar uchun yuqorida keltirilgan formuladan foydalanadi.
Yechim
Avval qisman kasr parchalanishining umumiy shaklini olaylik.
\[\frac{{{x^3} + 10{x^2} + 3x + 36}}{{\left( {x - 1} \right){{\left( {{x^2} + 4} \right)}^2}}}\, = \frac{A}{{x - 1}} + \frac{{Bx + C}}{{{x^2} + 4}} + \frac{{Dx + E}}{{{{\left( {{x^2} + 4} \right)}^2}}}\]Endi hisoblagichlarni teng qilib qo'ying, o'ng tomonni kengaytiring va o'xshash shartlarni to'plang.
\[\begin{align*}{x^3} + 10{x^2} + 3x + 36 & = A{\left( {{x^2} + 4} \right)^2} + \left( {Bx + C} \right)\left( {x - 1} \right)\left( {{x^2} + 4} \right) + \left( {Dx + E} \right)\left( {x - 1} \right)\\ & = \left( {A + B} \right){x^4} + \left( {C - B} \right){x^3} + \left( {8A + 4B - C + D} \right){x^2} + \\ & \hspace{0.5in}\hspace{0.25in}\left( { - 4B + 4C - D + E} \right)x + 16A - 4C - E\end{align*}\]Koeffitsientning tengligi quyidagi tizimni beradi.
\[\left. \begin{aligned}{x^4} & :\hspace{0.25in} & A + B & = 0\\ {x^3} & :\hspace{0.25in} & C - B & = 1\\ {x^2} & : \hspace{0.25in} & 8A + 4B - C + D & = 10\\ {x^1} & : \hspace{0.25in} & - 4B + 4C - D + E & = 3\\ {x^0} & :\hspace{0.25in} & 16A - 4C - E & = 36\end{aligned} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,A = 2,\,B = - 2,\,C = - 1,\,D = 1,\,E = 0\]Agar ba'zi koeffitsientlar nolga teng bo'lsa, hayajonlanmang. Vaqti-vaqti bilan sodir bo'ladi.
Mana integral.
\[\begin{align*}\int{{\frac{{{x^3} + 10{x^2} + 3x + 36}}{{\left( {x - 1} \right){{\left( {{x^2} + 4} \right)}^2}}}\,dx}} & = \int{{\frac{2}{{x - 1}} + \frac{{ - 2x - 1}}{{{x^2} + 4}} + \frac{x}{{{{\left( {{x^2} + 4} \right)}^2}}}\,dx}}\\ & = \int{{\frac{2}{{x - 1}} - \frac{{2x}}{{{x^2} + 4}} - \frac{1}{{{x^2} + 4}} + \frac{x}{{{{\left( {{x^2} + 4} \right)}^2}}}\,dx}}\\ & = 2\ln \left| {x - 1} \right| - \ln \left| {{x^2} + 4} \right| - \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) - \frac{1}{2}\frac{1}{{{x^2} + 4}} + c\end{align*}\]Shu paytgacha biz faqat ratsional iboralarni ko'rib chiqdik, bunda payning darajasi maxraj darajasidan qat'iy kam bo'lgan. Albatta, barcha ratsional iboralar bu shaklga mos kelmaydi va shuning uchun biz bunday bo'lmagan bir nechta misollarni ko'rib chiqishimiz kerak.
Yechim
Demak, bu holda ayiruvchining darajasi 4 ga, maxrajning darajasi esa 3 ga teng. Shuning uchun bu ratsional ifodada qisman kasrlar tuzish mumkin emas.
Buni tuzatish uchun biz buni hal qilishimiz mumkin bo'lgan shaklga keltirish uchun bu haqda uzoq bo'linishimiz kerak. Mana buning uchun ish.
\[\require{enclose} \begin{align*} &\,\,\, x - 2\\ {x^3} - 3{x^2} & \enclose{longdiv}{{x^4} - 5{x^3} + 6{x^2} - 18} \\ - & \underline {\left( {{x^4} - 3{x^3}} \right)\hspace{0.9in}} \\ & \hspace{0.3in} - 2{x^3} + 6{x^2} - 18\\ & \underline { \hspace{0.08in}-\left( { - 2{x^3} + 6{x^2}} \right) \hspace{0.4in}} \\ & \hspace{1.35in} - 18\end{align*}\]Shunday qilib, uzun bo'linishdan biz buni ko'ramiz,
\[\frac{{{x^4} - 5{x^3} + 6{x^2} - 18}}{{{x^3} - 3{x^2}}}\, = x - 2 - \frac{{18}}{{{x^3} - 3{x^2}}}\]va integral bo'ladi,
\[\begin{align*}\int{{\frac{{{x^4} - 5{x^3} + 6{x^2} - 18}}{{{x^3} - 3{x^2}}}\,dx}} & = \int{{x - 2 - \frac{{18}}{{{x^3} - 3{x^2}}}\,dx}}\\ & = \int{{x - 2\,dx}} - \int{{\frac{{18}}{{{x^3} - 3{x^2}}}\,dx}}\end{align*}\]Birinchi integralni biz yetarlicha oson bajara olamiz va ikkinchi integral endi qisman kasrlarni bajarishga imkon beruvchi shaklda. Shunday qilib, ikkinchi integrasiya uchun qisman kasrlarning umumiy shaklini olaylik.
\[\frac{{18}}{{{x^2}\left( {x - 3} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 3}}\]Numeratorlarni teng o'rnatish bizga beradi,
\[18 = Ax\left( {x - 3} \right) + B\left( {x - 3} \right) + C{x^2}\]Endi biz bu erda ishlaydigan birinchi ikkita misolda ishlatgan usulning o'zgarishi mavjud. \(x\) ning bir nechta qiymatlari mavjud bo'lib, ular bizga uchta doimiydan ikkitasini tezda olish imkonini beradi, ammo \(x\) ning hech qanday qiymati yo'q, u bizga faqat uchinchisini beradi.
Ushbu misolda biz osonlikcha olishimiz mumkin bo'lgan ikkita doimiyni olish uchun \(x\) ni tanlaymiz va keyin biz \(x\) ning boshqa qiymatini tanlaymiz, u bilan ishlash oson (i.e. u hech qanday joyda katta / tartibsiz raqamlarni bermaydi) va keyin biz uchinchisini topish uchun boshqa ikkita doimiyni ham bilishimizdan foydalanamiz.
\[\begin{align*}x & = 0 : & \hspace{0.25in} 18 & = B\left( { - 3} \right) & \hspace{0.15in}\Rightarrow \hspace{0.25in}B & = - 6\\ x & = 3 : & \hspace{0.25in} 18 & = C\left( 9 \right) & \hspace{0.15in} \Rightarrow \hspace{0.25in}C & = 2\\ x & = 1 : & 18 & = A\left( { - 2} \right) + B\left( { - 2} \right) + C = - 2A + 14 & \hspace{0.15in} \Rightarrow \hspace{0.25in}A & = - 2\end{align*}\]U holda integral,
\[\begin{align*}\int{{\frac{{{x^4} - 5{x^3} + 6{x^2} - 18}}{{{x^3} - 3{x^2}}}\,dx}} & = \int{{x - 2\,dx}} - \int{{ - \frac{2}{x} - \frac{6}{{{x^2}}} + \frac{2}{{x - 3}}\,dx}}\\ & = \frac{1}{2}{x^2} - 2x + 2\ln \left| x \right| - \frac{6}{x} - 2\ln \left| {x - 3} \right| + c\end{align*}\]Oldingi misolda maxrajda \({x^2}\) bilan ishlashning ikki xil usuli mavjud edi. Birinchisi, uni parchalanishda quyidagi atamani beradigan kvadratik sifatida ko'rib chiqishdir
\[\frac{{Ax + B}}{{{x^2}}}\]ikkinchisi esa uni chiziqli atama sifatida quyidagi tarzda ko'rib chiqishdir:
\[{x^2} = {\left( {x - 0} \right)^2}\]bu parchalanishda quyidagi ikkita atamani beradi,
\[\frac{A}{x} + \frac{B}{{{x^2}}}\]Biz misolimizda bu haqda ikkinchi fikrlash usulidan foydalandik. E'tibor bering, ikkalasi ham bir xil qisman parchalanishlarni beradi. Xo'sh, nega bu haqda gapirish kerak? Oddiy. Bu \({x^2}\) uchun ishlaydi, lekin \({x^3}\) yoki \({x^4}\) haqida nima deyish mumkin? Bunday hollarda, biz ushbu turdagi atamalar haqida ikkinchi fikrlash usulidan foydalanishimiz kerak bo'ladi.
\[{x^3}\,\,\, \Rightarrow \,\,\,\,\,\frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{{x^3}}}\hspace{0.5in}{x^4}\,\,\, \Rightarrow \,\,\,\,\,\frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{{x^3}}} + \frac{D}{{{x^4}}}\]Keling, yana bir misolni ko'rib chiqaylik.
Yechim
Bunda pay va maxraj bir xil darajaga ega. Oxirgi misolda bo'lgani kabi, buni to'g'ri shaklga keltirish uchun biz uzun bo'linishimiz kerak. Buning tafsilotlarini tekshirish uchun sizga qoldiramiz.
\[\int{{\frac{{{x^2}}}{{{x^2} - 1}}\,dx}} = \int{{1 + \frac{1}{{{x^2} - 1}}\,dx}} = \int{{dx}} + \int{{\frac{1}{{{x^2} - 1}}\,dx}}\]Shunday qilib, biz ikkinchi integralni qisman qismlarga ajratishimiz kerak. Mana parchalanish.
\[\frac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{x + 1}}\]Numeratorning tengligini o'rnatish,
\[1 = A\left( {x + 1} \right) + B\left( {x - 1} \right)\]\(x\) ni tanlash qiymati bizga quyidagi koeffitsientlarni beradi.
\[\begin{align*}x & = - 1 : & \hspace{0.25in} 1 & = B\left( { - 2} \right) & \hspace{0.25in} \Rightarrow \hspace{0.5in}B & = - \frac{1}{2}\\ x & = 1 \,\,\,\, : & \hspace{0.25in}1 & = A\left( 2 \right) & \hspace{0.25in} \Rightarrow \hspace{0.5in}A & = \frac{1}{2}\end{align*}\]U holda integral,
\[\begin{align*}\int{{\frac{{{x^2}}}{{{x^2} - 1}}\,dx}} & = \int{{dx}} + \int{{\frac{{\frac{1}{2}}}{{x - 1}} - \frac{{\frac{1}{2}}}{{x + 1}}\,dx}}\\ & = x + \frac{1}{2}\ln \left| {x - 1} \right| - \frac{1}{2}\ln \left| {x + 1} \right| + c\end{align*}\]