Keling, birinchi navbatda ildizni yo'q qilish uchun almashtirishdan foydalanamiz.

\[\sqrt {{x^2} + 16} = \sqrt {16{{\tan }^2}\left( \theta \right) + 16} = 4\sqrt {{{\sec }^2}\left( \theta \right)} = 4\left| {\sec \left( \theta \right)} \right|\]

Keyinchalik, biz noaniq integralni bajarayotganimiz uchun sekant musbat deb faraz qilamiz va shuning uchun biz olish uchun mutlaq qiymat satrlarini tashlab qo'yishimiz mumkin,

\[\sqrt {{x^2} + 16} = 4\sec \left( \theta \right)\]

O'zgartirishga tayyorgarlikning yakuniy bosqichi uchun, keling, differentsialni ham hisoblaylik, shuning uchun uni almashtirishda ishlatishni unutmang!

\[dx = 4{\sec ^2}\left( \theta \right)\,d\theta \]

Endi haqiqiy almashtirishni qilaylik.

\[\int{{\frac{{\sqrt {{x^2} + 16} }}{{{x^4}}}\,dx}} = \int{{\frac{{4\sec \left( \theta \right)}}{{{{\left( {4\tan \left( \theta \right)} \right)}^4}}}\,4{{\sec }^2}\left( \theta \right)d\theta }} = \int{{\frac{{{{\sec }^3}\left( \theta \right)}}{{16{{\tan }^4}\left( \theta \right)}}\,d\theta }}\]

Oldingi bosqichda biz hisoblagan differensialni almashtirishni unutmang. Bu, ehtimol, triglarni almashtirishda eng keng tarqalgan xatodir. Differensialni unutish muammoni sezilarli darajada o'zgartirishi mumkin, ko'pincha integralni baholash juda qiyin bo'ladi.

Endi biz integralni baholashimiz kerak. Bu holda integralni sekantlar va tangenslar nuqtai nazaridan bajarish biroz qiyin bo'lib tuyuladi, shuning uchun integralni sinuslar va kosinuslarga aylantiramiz va biz nimani olishimizni ko'rib chiqamiz. Shunday qilib,

\[\int{{\frac{{\sqrt {{x^2} + 16} }}{{{x^4}}}\,dx}} = \frac{1}{{16}}\int{{\frac{{\cos \left( \theta \right)}}{{{{\sin }^4}\left( \theta \right)}}\,d\theta }}\]

Bu baholash uchun oddiy integral, shuning uchun bu erda integral baholash.

\[\begin{align*}\int{{\frac{{\sqrt {{x^2} + 16} }}{{{x^4}}}\,dx}} & = \frac{1}{{16}}\int{{\frac{{\cos \left( \theta \right)}}{{{{\sin }^4}\left( \theta \right)}}\,d\theta }}\hspace{0.25in}\hspace{0.25in}u = \sin \left( \theta \right)\\ & = \frac{1}{{16}}\int{{{u^{ - 4}}\,du}}\\ & = - \frac{1}{{48}}{u^{ - 3}} + c = - \frac{1}{{48}}{\left[ {\sin \left( \theta \right)} \right]^{ - 3}} + c = - \frac{1}{{48}}{\csc ^3}\left( \theta \right) + c\end{align*}\]

Trig funksiyalarini o'z ichiga olgan integrallarni baholash uchun biz tez-tez bajarishimiz kerak bo'lgan integralning barcha "standart" manipulyatsiyalarini unutmang. Agar siz ularni eslamasangiz, oldingi bo'limga qaytib, ularni yaxshi o'rganish uchun ba'zi amaliy muammolarni hal qilishingiz kerak bo'ladi.

Har bir trigni almashtirish muammosi trig funksiyalarini o'z ichiga olgan integralgacha kamayadi va ularning ko'pchiligini baholash uchun integralni biroz manipulyatsiya qilish kerak bo'ladi.

Yakuniy qadam sifatida biz \(x\)-ga qaytishimiz kerak. Buning uchun bizga tez uchburchak kerak bo'ladi. Mana o'sha ish.

Bizda mavjud almashtirishdan, \[\tan \left( \theta \right) = \frac{x}{4}\,\,\,\,\,\left( { = \frac{{{\mbox{opp}}}}{{{\mbox{adj}}}}} \right)\] To'g'ri uchburchakdan biz olamiz, \[\csc \left( \theta \right) = \frac{{\sqrt {{x^2} + 16} }}{x}\]

U holda integral,

\[\int{{\frac{{\sqrt {{x^2} + 16} }}{{{x^4}}}\,dx}} = - \frac{1}{{48}}{\left[ {\frac{{\sqrt {{x^2} + 16} }}{x}} \right]^3} + c = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{{{\left( {{x^2} + 16} \right)}^{\frac{3}{2}}}}}{{48{x^3}}} + c}}\]