5.4-bo'lim : Ko'proq almashtirish qoidasi
11. \( \displaystyle \int{{\frac{{8 - w}}{{4{w^2} + 9}}\,dw}}\) ni baholang.
Maslahat: Bu erda yozilgan integrand bilan bu muammoni hal qilib bo'lmaydi.
Yechim
Aytganimizdek, biz bu muammoni hal qila olmaymiz. Ushbu integralni bajarish uchun biz integralni quyidagi tarzda qayta yozishimiz kerak.
\[\int{{\frac{{8 - w}}{{4{w^2} + 9}}\,dw}} = \int{{\frac{8}{{4{w^2} + 9}}\,dw}} - \int{{\frac{w}{{4{w^2} + 9}}\,dw}}\]2-qadam
Endi birinchi integral teskari tangens bo'lishi mumkin (garchi biz bu integralni qayta yozishimiz kerak bo'lsa-da), ikkinchisi esa logarifmaga o'xshaydi (tez almashtirish bilan).
Shunday qilib, birinchi integralni qayta yozish.
\[\int{{\frac{{8 - w}}{{4{w^2} + 9}}\,dw}} = \frac{8}{9}\int{{\frac{1}{{\frac{4}{9}{w^2} + 1}}\,dw}} - \int{{\frac{w}{{4{w^2} + 9}}\,dw}}\]3-qadam
Endi bizga har bir integral uchun almashtirish kerak bo'ladi. Bu erda har bir integral uchun kerak bo'ladigan almashtirishlar.
\[u = \frac{2}{3}w\,\,\,\,\,\,\left( {{\mbox{so }}{u^2} = \frac{4}{9}{w^2}} \right)\hspace{0.75in}v = 4{w^2} + 9\]4-qadam
Bu erda almashtirish uchun differentsial ish.
\[du = \frac{2}{3}dw\hspace{0.25in} \to \hspace{0.25in}dw = \frac{3}{2}du\hspace{0.5in}dv = 8w\,dw\,\,\,\,\,\,\, \to \hspace{0.25in}\,w\,dw = \frac{1}{8}dv\]Endi almashtirishlarni bajarish va integrallarni baholash,
\[\begin{align*}\int{{\frac{{8 - w}}{{4{w^2} + 9}}\,dw}} & = \frac{8}{9}\left( {\frac{3}{2}} \right)\int{{\frac{1}{{{u^2} + 1}}\,du}} - \frac{1}{8}\int{{\frac{1}{v}\,dv}} = \frac{4}{3}{\tan ^{ - 1}}\left( u \right) - \frac{1}{8}\ln \left| v \right| + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{4}{3}{{\tan }^{ - 1}}\left( {\frac{2}{3}w} \right) - \frac{1}{8}\ln \left| {4{w^2} + 9} \right| + c}}\end{align*}\]Integralni baholagandan so'ng asl o'zgaruvchiga qaytishni unutmang!