integral
Posted: 24/10-2007 12:54
Lurer på om man kan gjøre det sånn.
[tex]I=\int \frac{dx}{x^2 \sqrt{x^2 +1}}[/tex]
[tex]x=tan\theta[/tex]
[tex]dx=sec^2 \theta d\theta[/tex]
[tex]I=\int \frac{sec^2 \theta d\theta}{tan^2 \theta \sqrt{sec^2 \theta}}[/tex]
[tex]I=\int \frac{sec \theta d\theta}{tan^2 \theta}[/tex]
[tex]I=\int \frac{\frac{1}{cos\theta} d\theta}{ \frac{sin^2 \theta}{cos^2 \theta}}[/tex]
[tex]I=\int \frac{d\theta}{ \frac{sin^2 \theta}{cos \theta}}[/tex]
[tex]I=\int \frac{d\theta}{ \frac{sin^2 \theta}{cos \theta}}[/tex]
[tex]I=\int \frac{cos\theta d\theta}{sin^2 \theta}[/tex]
[tex]u=sin \theta[/tex]
[tex]du=cos \theta[/tex]
[tex]I=\int \frac{du}{u^2} = -\frac{1}{u} + C = -\frac{1}{sin\theta} + C[/tex] = [tex] -\frac{1}{\frac{x}{\sqrt{x^2 +1}}} + C[/tex] = [tex] -\frac{\sqrt{x^2 +1}}{x} + C[/tex]
[tex]I=\int \frac{dx}{x^2 \sqrt{x^2 +1}}[/tex]
[tex]x=tan\theta[/tex]
[tex]dx=sec^2 \theta d\theta[/tex]
[tex]I=\int \frac{sec^2 \theta d\theta}{tan^2 \theta \sqrt{sec^2 \theta}}[/tex]
[tex]I=\int \frac{sec \theta d\theta}{tan^2 \theta}[/tex]
[tex]I=\int \frac{\frac{1}{cos\theta} d\theta}{ \frac{sin^2 \theta}{cos^2 \theta}}[/tex]
[tex]I=\int \frac{d\theta}{ \frac{sin^2 \theta}{cos \theta}}[/tex]
[tex]I=\int \frac{d\theta}{ \frac{sin^2 \theta}{cos \theta}}[/tex]
[tex]I=\int \frac{cos\theta d\theta}{sin^2 \theta}[/tex]
[tex]u=sin \theta[/tex]
[tex]du=cos \theta[/tex]
[tex]I=\int \frac{du}{u^2} = -\frac{1}{u} + C = -\frac{1}{sin\theta} + C[/tex] = [tex] -\frac{1}{\frac{x}{\sqrt{x^2 +1}}} + C[/tex] = [tex] -\frac{\sqrt{x^2 +1}}{x} + C[/tex]