R2 2013 vår LØSNING
DEL EN
Oppgave 1.
a) <math> \displaystyle f'(x) = - 3 \cos x </math>
b) <math> \displaystyle g'(x) = 6\pi \cos(\pi x) </math>
c) <math> \displaystyle h'(x) = \left[ 2 \sin(3x) + 3 \right] \cos(3x) e^{2x} </math>
Oppgave 2.
a) La $u = 2x$ da er $\mathrm{d}u = 2x \mathrm{d}x$ slik at
$ \displaystyle \int \frac{2x}{x^2-4}\,\mathrm{d}x = \int \frac{\mathrm{d}u}{u} = \ln|u| +\mathcal{C} = \ln \left| x^2 - 4 \right| +\mathcal{C}$
b) Legg merke til at
$\displaystyle \frac{2x}{x^2-4} = \frac{x + x}{(x-2)(x+2)} = \frac{(x-2) + (x+2)}{(x-2)(x+2)} = \frac{(x-2)}{(x-2)(x+2)} + \frac{(x+2)}{(x-2)(x+2)} = \frac{1}{x+2} + \frac{1}{x-2} $
slik at vi kan skrive integralet som
$ \displaystyle \int \frac{2x}{x^2-4}\,\mathrm{d}x = \int \frac{1}{x+2} + \frac{1}{x-2} \, \mathrm{d}x = \ln\left| x + 2 \right| + \ln\left| x - 2\right| + \mathcal{C} = \ln \left| x^2 - 4 \right| +\mathcal{C} $
som ønsket.