Интеграл...

[beatjunky]

Да се пресметне интеграла...
1
[tex]\int_{}^{ } \frac{\sqrt{x}}{ 2x+1}.dx[/tex]
0

[martin123456]

полагаме [tex]\sqrt{x}=y[/tex]
интегралът става [tex]\int_{0}^{1}\frac{ydy^2}{2y^2+1}=\int_{0}^{1}{\frac{2y^2dy}{2y^2+1}}=\int_{0}^{1}dy-\int_{0}^{1}{\frac{dy}{2y^2+1}}=1-0-\frac{1}{\sqrt{2}}\int_{0}^{1}{\frac{d(\sqrt{2}y)}{(\sqrt{2}y)^2+1}}=1-\frac{1}{\sqrt{2}}\int_{0}^{\frac{1}{\sqrt{2}}}{\frac{dz}{z^2+1}}=1-\frac{1}{\sqrt{2}}arctg|_{0}^{\frac{1}{\sqrt{2}}}=[/tex][tex]1-\frac{1}{\sqrt{2}}arctg\frac{1}{\sqrt{2}}[/tex]
май...

[asdqwezxc]

[tex]1 - \frac{ arctg - (\sqrt{2} ) } { \sqrt{2} }[/tex]