Някой ще може ли да ми реши тези интеграли

[Sasho0o]

Някой ще може ли да ми реши тези интеграли

∫3√(1-2x)dx

∫(x+3)/(x2+1)dx

∫1/(2-x2)dx

[Реклама]

[martin123456]

1)[tex]\int{3\sqrt{1-2x}dx}=\frac{3}{2}\int{\sqrt{1-2x}d2x=-\frac{3}{2}\int{\sqrt{1-2x}d(1-2x)=-\frac{3}{2}\int{(1-2x)^{\frac{1}{2}}d(1-2x)}=-\frac{3}{2}(1-2x)^{\frac{3}{2}}\frac{2}{3}=-(1-2x)^{\frac{3}{2}}[/tex]

[martin123456]

[tex]\int{\frac{x+3}{x^2+1}dx}=\int{\frac{x}{x^2+1}dx}+3\int{\frac{1}{x^2+1}dx}=\frac{1}{2}\int{\frac{d(x^2+1)}{x^2+1}}+3arctgx=\frac{1}{2}ln(x^2+1)+3arctgx[/tex]

[martin123456]

[tex]\int{\frac{1}{2-x^2}dx}=\int{\frac{\frac{1}{2}}{1-(\frac{x}{\sqrt{2}})^2}dx=\frac{\sqrt{2}}{2}\int{\frac{1}{1-y^2}dy}[/tex], където сме положили [tex]y=\frac{x}{\sqrt{2}}[/tex]
[tex]\frac{1}{1-y^2}=\frac{1}{(y+1)(y-1)}=\frac{1}{2}(-\frac{1}{y+1}+\frac{1}{y-1})[/tex]
[tex]\int{\frac{1}{1-y^2}dy=\frac{1}{2}(-ln|y+1|+ln|1-y|)[/tex]