Nok et integral...[løst]
Lagt inn: 26/04-2008 14:04
Har stykket:
[tex]\int \frac{2x+4}{x^2+4x+3}dx[/tex]
[tex]= \int \frac{2x+4}{(x+1)(x+5)}dx[/tex]
[tex]2x+4 = A(x+5) + B(x+1)[/tex]
Setter x=-5
[tex]-6 = -4B[/tex], [tex]B=\frac{3}{2}[/tex]
Setter x = -1
[tex]2 = 4A[/tex], [tex]A = \frac{1}{2}[/tex]
[tex]\int \frac{2x+4}{x^2+4x+3}dx = \int \frac{\frac{1}{2}}{(x+1)} + \frac{\frac{3}{2}}{(x+5)}dx[/tex]
[tex] = \frac{1}{2}ln |x+1| + \frac{3}{2}ln|x+5| + C[/tex]
Men fasit sier:
[tex]ln|x^2 + 4x + 3| =[/tex] [tex]ln|x+3|[/tex] + [tex]ln|x+1|[/tex] + [tex]C[/tex]
Hvor tryner jeg???
[tex]\int \frac{2x+4}{x^2+4x+3}dx[/tex]
[tex]= \int \frac{2x+4}{(x+1)(x+5)}dx[/tex]
[tex]2x+4 = A(x+5) + B(x+1)[/tex]
Setter x=-5
[tex]-6 = -4B[/tex], [tex]B=\frac{3}{2}[/tex]
Setter x = -1
[tex]2 = 4A[/tex], [tex]A = \frac{1}{2}[/tex]
[tex]\int \frac{2x+4}{x^2+4x+3}dx = \int \frac{\frac{1}{2}}{(x+1)} + \frac{\frac{3}{2}}{(x+5)}dx[/tex]
[tex] = \frac{1}{2}ln |x+1| + \frac{3}{2}ln|x+5| + C[/tex]
Men fasit sier:
[tex]ln|x^2 + 4x + 3| =[/tex] [tex]ln|x+3|[/tex] + [tex]ln|x+1|[/tex] + [tex]C[/tex]
Hvor tryner jeg???