7.6-bo'lim : Kvadratlarni o'z ichiga olgan integrallar
2. \( \displaystyle \int{{\frac{{10x}}{{4{x^2} - 8x + 9}}\,dx}}\) integralini baholang.
Yechim
Birinchi narsa qilish kerak bo'lgan kvadratni maxrajdagi kvadratchada to'ldirish (kvadrat tafsilotlarini to'ldirishni tekshirishni sizga qoldiramiz).
\[\int{{\frac{{10x}}{{4{x^2} - 8x + 9}}\,dx}} = \int{{\frac{{10x}}{{4{{\left( {x - 1} \right)}^2} + 5}}\,dx}}\]2-qadam
Bundan ko'rishimiz mumkinki, quyidagi almashtirish biz uchun ishlashi kerak.
\[u = x - 1\hspace{0.5in} \Rightarrow \hspace{0.5in}du = dx\hspace{0.25in}\& \hspace{0.25in}x = u + 1\]O'zgartirishni amalga oshirish beradi,
\[\int{{\frac{{10x}}{{4{x^2} - 8x + 9}}\,dx}} = \int{{\frac{{10\left( {u + 1} \right)}}{{4{u^2} + 5}}\,du}}\]3-qadam
Agar biz uni quyidagi tarzda ajratsak, bu integralni tezda bajarishimiz mumkin,
\[\int{{\frac{{10x}}{{4{x^2} - 8x + 9}}\,dx}} = \int{{\frac{{10u}}{{4{u^2} + 5}}\,du}} + \int{{\frac{{10}}{{4{u^2} + 5}}\,du}} = \int{{\frac{{10u}}{{4{u^2} + 5}}\,du}} + \frac{5}{2}\int{{\frac{1}{{{u^2} + \frac{5}{4}}}\,du}}\]Ikkinchi integralni tezda qayta yozgandan so'ng, biz birinchisini \(v = 4{u^2} + 5\) almashtirish bilan amalga oshirishimiz mumkinligini ko'ramiz, ikkinchisi esa teskari trig integral bo'lib, biz ushbu bo'lim uchun eslatmalarning boshida berilgan formuladan foydalanib baholashimiz mumkin.
\[\begin{align*}\int{{\frac{{10x}}{{4{x^2} - 8x + 9}}\,dx}} &= \frac{5}{4}\ln \left| v \right| + \frac{5}{2}\left( {\frac{2}{{\sqrt 5 }}} \right){\tan ^{ - 1}}\left( {\frac{{2u}}{{\sqrt 5 }}} \right) + c\\ & = \frac{5}{4}\ln \left| {4{u^2} + 5} \right| + \sqrt 5 {\tan ^{ - 1}}\left( {\frac{{2u}}{{\sqrt 5 }}} \right) + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{5}{4}\ln \left| {4{{\left( {x - 1} \right)}^2} + 5} \right| + \sqrt 5 {{\tan }^{ - 1}}\left( {\frac{{2x - 2}}{{\sqrt 5 }}} \right) + c}}\end{align*}\]\(u\) o'rniga qayta yozishni unutmang!