Uppgift 3

(a) En linjär substitution ger
$$\begin{aligned}\int\frac x{\sqrt{x-2}}\,dx&=\left[\begin{array}{c}t=x-2\\dt=dx\end{array}\right]=\int\frac{t+2}{\sqrt t}\,dt=\int(t^{1/2}+2t^{-1/2})\,dt\\&=\frac23t^{3/2}+4t^{1/2}+C=t^{1/2}\left(\frac{2t}3+4\right)+C=\frac23(x+4)\sqrt{x-2}+C.\end{aligned}$$
(b) Vi får
$$\int\frac{\arctan(3x)}{1+9x^2}\,dx=\left[\begin{array}{c}u=\arctan(3x)\\du=\frac3{1+9x^2}\,dx\end{array}\right]=\frac13\int u\,du=\frac{u^2}6+C=\frac{\arctan(3x)^2}6+C,$$
så $\int_0^\infty\frac{\arctan(3x)}{1+9x^2}\,dx=\lim_{N\to\infty}\left[\frac{\arctan(3x)^2}6\right]_0^N=\frac16\left(\frac\pi2\right)^2=\frac{\pi^2}{24}$.