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Løs følgende vgs-integral:

[tex]I=\int_{0}^{\sqrt{3}}\frac{1}{1-x^4}\,d(3x-x^3)[/tex]

Janhaa skrev:Løs følgende vgs-integral:

[tex]I=\int_{0}^{\sqrt{3}}\frac{1}{1-x^4}\,d(3x-x^3)[/tex]

sant [tex]dx\neq d(3x-x^3)[/tex]


[tex]\int \frac{1}{1-x^4}d(3x-x^3)=\int \frac{1}{1-x^4}dx(3-x^2)=\int \frac{(3-x^2)}{1-x^4}dx[/tex]


[tex]\int \frac{(3-x^2)}{1-x^4}dx=\int \frac{-x^2+3}{-x^4+1}dx=\int \frac{x^2-3}{\left ( x^2+1 \right )\left ( x-1 \right )\left ( x+1 \right )}[/tex]


delbrøkoppspaltning:

[tex]\frac{x^2-3}{(x^2+1)(x-1)(x+1)}=\frac{A}{x^2+1}+\frac{B}{x-1}+\frac{C}{x+1}[/tex]

får da ;[tex]\frac{x^2-3}{(x^2+1)(x-1)(x+1)}=\frac{2}{x^2+1}+\frac{\frac{1}{2}}{x+1}+\frac{-\frac{1}{2}}{x-1}[/tex]


[tex]\int \frac{x^2-3}{x^4-1}dx=\frac{1}{2}\int \frac{1}{x+1}dx-\frac{1}{2}\int \frac{1}{x-1}dx+2\int \frac{1}{x^2+1}dx[/tex]

* [tex]\frac{1}{2}\int \frac{1}{x+1}=\frac{1}{2}\ln\left | x+1 \right |+C[/tex]

* [tex]-\frac{1}{2}\int \frac{1}{x-1}dx=-\frac{1}{2}\ln \left | x-1 \right |+C[/tex]

* [tex]2\int \frac{1}{x^2+1}dx=?[/tex]

Har ikke sett dette integralet før
massive fail..

Drezky skrev:

Janhaa skrev:Løs følgende vgs-integral:

[tex]I=\int_{0}^{\sqrt{3}}\frac{1}{1-x^4}\,d(3x-x^3)[/tex]

sant [tex]dx\neq d(3x-x^3)[/tex]

Du må la $u=3x-x^3$. Fra kjerneregelen blir dermed $d(3x-x^3)=du=\frac{du}{dx}dx=(3-3x^2)dx$

Drezky skrev:

Janhaa skrev:Løs følgende vgs-integral:
[tex]I=\int_{0}^{\sqrt{3}}\frac{1}{1-x^4}\,d(3x-x^3)[/tex]

sant [tex]dx\neq d(3x-x^3)[/tex]
[tex]\int \frac{1}{1-x^4}d(3x-x^3)=\int \frac{1}{1-x^4}dx(3-x^2)=\int \frac{(3-x^2)}{1-x^4}dx[/tex]
[tex]\int \frac{(3-x^2)}{1-x^4}dx=\int \frac{-x^2+3}{-x^4+1}dx=\int \frac{x^2-3}{\left ( x^2+1 \right )\left ( x-1 \right )\left ( x+1 \right )}[/tex]
* [tex]\frac{1}{2}\int \frac{1}{x+1}=\frac{1}{2}\ln\left | x+1 \right |+C[/tex]
* [tex]-\frac{1}{2}\int \frac{1}{x-1}dx=-\frac{1}{2}\ln \left | x-1 \right |+C[/tex]
* [tex]2\int \frac{1}{x^2+1}dx=?[/tex]
Har ikke sett dette integralet før
massive fail..

Litt annerledes integrasjon ja...den blir noe sånt, med info fra plutarco:

[tex]I=3\int_0^{\sqrt{3}}\frac{(1-x^2)\,dx}{(1-x^2)(1+x^2)}=3(\arctan(\sqrt{3}) - \arctan(0)) = \pi[/tex]