Bunday holda, almashtirish kerak kabi ko'rinadi

\[u = 1 + 9{x^2}\]

Mana bu almashtirish uchun farq.

\[du = 18x\,dx\hspace{0.5in} \Rightarrow \hspace{0.5in}3x\,dx = \frac{1}{6}du\]

U holda integral,

\[\int{{\frac{{3x}}{{1 + 9{x^2}}}\,dx}} = \frac{1}{6}\int{{\frac{1}{u}\,du}} = \frac{1}{6}\ln \left| u \right| + c = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{6}\ln \left| {1 + 9{x^2}} \right| + c}}\]

Bu qism uchun almashtirish va differensial ish oldingi qism bilan bir xil.

\[u = 1 + 9{x^2}\hspace{0.5in}du = 18x\,dx\hspace{0.5in} \Rightarrow \hspace{0.5in}\,3x\,dx = \frac{1}{6}du\]

Mana bu qism uchun integral,

\[\int{{\frac{{3x}}{{{{\left( {1 + 9{x^2}} \right)}^4}}}\,dx}} = \frac{1}{6}\int{{\frac{1}{{{u^4}}}\,du}} = \frac{1}{6}\int{{{u^{ - 4}}\,du}} = - \frac{1}{{18}}{u^{ - 3}} + c = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{1}{{18}}\frac{1}{{{{\left( {1 + 9{x^2}} \right)}^3}}} + c}}\]

1/(bir narsa) ko‘rinishidagi funksiyalarning har bir integralini logarifmlarga aylantirmaslikdan ehtiyot bo‘ling! Bu talabalar tez-tez uchraydigan xatolardan biridir.

Bizda endi numeratorda \(x\) mavjud emasligi sababli, bu integral oldingi ikkitasidan juda farq qiladi. O'zgartirish aniqroq bo'lishi uchun integralni tezda qayta yozamiz.

\[\int{{\frac{3}{{1 + 9{x^2}}}\,dx}} = \int{{\frac{3}{{1 + {{\left( {3x} \right)}^2}}}\,dx}}\]

Demak, bu almashtirishni talab qiladigan teskari tangens muammosiga o'xshaydi.

\[u = 3x\hspace{0.5in} \to \hspace{0.5in}du = 3dx\]

U holda integral,

\[\int{{\frac{3}{{1 + 9{x^2}}}\,dx}} = \int{{\frac{1}{{1 + {u^2}}}\,du}} = {\tan ^{ - 1}}\left( u \right) + c = \require{bbox} \bbox[2pt,border:1px solid black]{{{{\tan }^{ - 1}}\left( {3x} \right) + c}}\]