Pent som vanlig Jarle.Jarle10 skrev:[tex]I_6=\int \frac{x^2-1}{x(x^2+1)sqrt{x^2+\frac{1}{x^2}}}\, \rm{d}x=\int \frac{x^2-1}{x(x^2+1)sqrt{(x+\frac{1}{x})^2-2}}\, \rm{d}x[/tex]
[tex]\sqrt{2}u=x+\frac{1}{x} \Rightarrow \sqrt{2}\frac{\rm{d}u}{\rm{d}x}=1-\frac{1}{x^2} = \frac{1}{x^2}(x^2-1) \Rightarrow \sqrt{2}x^2\frac{\rm{d}u}{\rm{d}x}=x^2-1[/tex]
[tex]\sqrt{2}ux=x^2+1[/tex]
[tex]I_6= \int \frac{\sqrt{2}x^2}{x(\sqrt{2}ux)\sqrt{2u^2-2}} \rm{d}u = \frac{\sqrt{2}}{2} \int \frac{1}{u\sqrt{u^2-1}} \rm{d}u[/tex]
[tex]u^2-1=t \Rightarrow \frac{\rm{d}t}{\rm{d}u}=2u[/tex]
[tex]I_6=\frac{\sqrt{2}}{4} \int \frac{1}{u^2\sqrt{t}} \rm{d}t=\frac{\sqrt{2}}{4} \int \frac{1}{(t+1)\sqrt{t}} \rm{d}t[/tex]
[tex]t=r^2 \Rightarrow 2r\frac{\rm{d}r}{\rm{d}t} = 1[/tex]
[tex]I_6=\frac{\sqrt{2}}{4} \int \frac{2r}{(r^2+1)r} \rm{d}r=\frac{\sqrt{2}}{2} \int \frac{1}{(r^2+1)} \rm{d}r=\frac{\sqrt{2}}{2}\arctan(r)+C[/tex]
[tex]r=\sqrt{t}=\sqrt{u^2-1}=\sqrt{\frac{1}{2}(x+\frac{1}{x})^2-1}=\sqrt{\frac{1}{2}(x^2+\frac{1}{x^2})}[/tex]
[tex]I_6=\frac{\sqrt{2}}{2}\arctan(\sqrt{\frac{1}{2}(x^2+\frac{1}{x^2})})+C[/tex]
Hvis noen sitter å venter med et integral, bare post det. Hvis ikke kan jeg prøve å spore opp et selv.
Ok, nytt integral:
[tex]I_7=\int \frac{(2-x)^{2n-1}}{(2+x)^{2n+1}}\,dx\,\,\,\,\,\,\text for n \geq 1[/tex]