2 интеграла - x^5lnx

[L.e.o]

[tex]\int_{}^{ }x^5 lnx dx = ?[/tex]
[tex]\int_{}^{ } \sqrt{9-x^2}dx = ?[/tex]

[ferry2]

[tex]\int_{}^{ }x^5 lnx dx = \int \ln xd\left(\frac{x^6}{6}\right)=\frac{x^6}{6}\ln x - \frac{1}{6}\int x^5 d(\ln x)=\frac{x^6}{6}\ln
x - \frac{1}{6}\int x^4=\frac{x^6}{6}\ln x - \frac{x^5}{30}[/tex]

[tex]\int_{}^{ } \sqrt{9-x^2}dx[/tex] Полагаме [tex]x=3\sin u, dx= 3\cos u du[/tex]

[tex]\Rightarrow \int_{}^{ } \sqrt{9-x^2}dx = \int \sqrt{9-9\sin ^2x} \cdot 3\cos udu = 9\int \cos ^2udu = \ldots[/tex]

[L.e.o]

Благодаря.

[kerry]

[tex]\int_{}^{ }x^5 lnx dx = \int \ln xd\left(\frac{x^6}{6}\right)=\frac{x^6}{6}\ln x - \frac{1}{6}\int x^5 d(\ln x)=\frac{x^6}{6}\ln
x - \frac{1}{6}\int x^4=\frac{x^6}{6}\ln x - \frac{x^5}{30}[/tex]

[tex]\int_{}^{ }x^5 lnx dx = \int \ln xd\left(\frac{x^6}{6}\right)=\frac{x^6}{6}\ln x - \frac{1}{6}\int x^6 d(\ln x)=....[/tex]