Uppgift 5

Vi gör variabelsubstitutionen \(u = e^x\), detta ger \(du = e^x dx\) och integrationsgränserna \(u(0) = 1\) samt \(u(\ln \sqrt{3}) = \sqrt{3}\).

$$\int_1^{\sqrt{3}} \frac{du}{3u^2 + 1} = \int_1^{\sqrt{3}} \frac{du}{(\sqrt{3}u)^2 + 1^2}$$

$$= \left( \frac{arctan(\sqrt{3}u)}{\sqrt{3}} \right)_1^{\sqrt{3}}$$

$$= \frac{1}{\sqrt{3}}\left( \arctan(3) - \arctan(\sqrt{3}) \right)$$

$$= \frac{1}{\sqrt{3}}\left( \arctan(3) - \frac{\pi}{3} \right)$$