Am urmatoarea integrala:
![\[ \int\limits_0^1 {\frac{{\sqrt x }}{{1 + \sqrt[3]{x}}}} dx \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-044704d5d892c3ba9b653faf2659196a_l3.png "Rendered by QuickLaTeX.com")
Trebuie calculata cu formula Leibniz-Newton, ceea ce nu constituie o problema. Dificultatea apare la calculul primitivei.
Am inteles ca in astfel de cazuri, se face o substitutie de forma
![\[ x = u^{cmmmc\left( {n,p} \right)} \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-b2e2ed2402bead0971e230e17d34e54a_l3.png "Rendered by QuickLaTeX.com")
unde n si p sunt indicii radicalilor. Evident,
![\[ n \ne p \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-06aba2c9bd34af013ed36ebc2720725e_l3.png "Rendered by QuickLaTeX.com")
. In cazul nostru avem 
deci
![\[ dx = 6u^5 du \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-834b0ad48faf3545ec6140b9dcaae818_l3.png "Rendered by QuickLaTeX.com")
Inlocuim, si avem:
![\[ \int {\frac{{\sqrt x }}{{1 + \sqrt[3]{x}}}dx} = \int {\frac{{u^3 }}{{1 + u^2 }}6u^5 du = 6\int {\frac{{u^8 }}{{1 + u^2 }}du} } \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-b9d740f10a291525bf0b618fd7d092e0_l3.png "Rendered by QuickLaTeX.com")
Am incercat sa calculez prin parti aceasta integrala:
![\[ \begin{array}{l} h\left( u \right) = u^8 \Rightarrow h'\left( u \right) = 8u^7 \\ v'\left( u \right) = \frac{1}{{1 + u^2 }} \Rightarrow v\left( u \right) = arctg\left( u \right) \\ \int {\frac{{u^8 }}{{1 + u^2 }}du} = u^8 arctg\left( u \right) - 8\int {u^7 arctg\left( u \right)} \\ \end{array} \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-8adb1af8cfb4bda56b615eea75a35f98_l3.png "Rendered by QuickLaTeX.com")
Acum am incercat, de asemenea, sa calculez ultima integrala tot prin parti:
![\[ \begin{array}{l} h\left( u \right) = arctg\left( u \right) \Rightarrow h'\left( u \right) = \frac{1}{{1 + u^2 }} \\ v'\left( u \right) = u^7 \Rightarrow v\left( u \right) = \frac{{u^8 }}{8} \\ \int {u^7 arctg\left( u \right)} = \frac{{u^8 }}{8}arctg\left( u \right) - \frac{1}{8}\int {\frac{{u^8 }}{{1 + u^2 }}} \\ \end{array} \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-4dbcb1e0df0451fc4c866281feae33ec_l3.png "Rendered by QuickLaTeX.com")
Acum introduc integrala in prima aplicare a formulei de integrare prin parti:
![\[ \int {\frac{{u^8 }}{{1 + u^2 }}du} = u^8 arctg\left( u \right) - 8\left( {\frac{{u^8 }}{8}arctg\left( u \right) - \frac{1}{8}\int {\frac{{u^8 }}{{1 + u^2 }}} } \right) = u^8 arctg\left( u \right) - u^8 arctg\left( u \right) + \int {\frac{{u^8 }}{{1 + u^2 }}} \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-6b5f1a94822d0c50658f9bc0694d4d2d_l3.png "Rendered by QuickLaTeX.com")
Care conduce, din cate se vede, la o anomalie.
Cum se rezolva? E corecta substitutia?
Salut! Prima substitutie e corecta!
La acea ultima integrala trebuie sa faci impartirea! Gradul numaratorului e mai mare decat gradul numitorului iar regula de integrare a functiilor rationale spune ca daca gradul numaratorului e mai mare decat gradul numitorului atunci se face impartirea!
trebuie sa o aduci la forma :
S(au^6 +bu^4+cu^2+d)du + k*S(du)/(1+u^2)=…
a,b,c,d,k – coeficienti rezultati in urma impartirii
Am reusit! Multumesc mult!