7.5-bo'lim : Ildizlarni o'z ichiga olgan integrallar
1. \( \displaystyle \int{{\frac{7}{{2 + \sqrt {x - 4} }}\,dx}}\) integralini baholang.
Yechim
Bu erda biz foydalanadigan almashtirish:
\[u = \sqrt {x - 4} \]2-qadam
Endi biz almashtirishga tayyorlanishimiz kerak. Boshqacha qilib aytganda, biz \(x\) ni hal qilishimiz va \(dx\) ni olishimiz kerak.
\[x = {u^2} + 4\hspace{0.5in} \Rightarrow \hspace{0.5in}dx = 2u\,du\]3-qadam
O'zgartirishni amalga oshirish beradi,
\[\int{{\frac{7}{{2 + \sqrt {x - 4} }}\,dx}} = \int{{\frac{7}{{2 + u}}\,\left( {2u} \right)du}} = \int{{\frac{{14u}}{{2 + u}}\,du}}\]4-qadam
Ushbu yangi integral \(v = u + 2\) almashtirish bilan amalga oshirilishi mumkin. Shunday qilib,
\[\int{{\frac{7}{{2 + \sqrt {x - 4} }}\,dx}} = \int{{\frac{{14\left( {v - 2} \right)}}{v}\,dv}} = \int{{14 - \frac{{28}}{v}\,dv}} = 14v - 28\ln \left| v \right| + c\]5-qadam
Oxirgi qadam, yakuniy javobni olish uchun barcha orqadagi almashtirishlarni bajarishdir.
\[\int{{\frac{7}{{2 + \sqrt {x - 4} }}\,dx}} = 14\left( {u + 2} \right) - 28\ln \left| {u + 2} \right| + c = \require{bbox} \bbox[2pt,border:1px solid black]{{14\left( {\sqrt {x - 4} + 2} \right) - 28\ln \left| {\sqrt {x - 4} + 2} \right| + c}}\]E'tibor bering, agar biz dastlabki almashtirish uchun \(u = \sqrt {x - 4} + 2\) dan foydalansak, ikkinchi almashtirishdan qochishimiz mumkin edi.
Bu ko'pincha ishlamaydi, lekin bu holda maxrajdagi yagona qo'shimcha atama doimiy bo'lgani uchun u differentsial ishni o'zgartirmadi va shuning uchun bu muammo uchun juda oson ishlaydi.