Uppgift 3

(a) Med partialintegrering får vi $\int_1^e x^2\ln(x)\,dx=\left[\frac{x^3}3\ln(x)\right]_1^e-\int_1^e\frac{x^3}3\frac1x\,dx=\frac{e^3}3-\frac13\int_1^e x^2\,dx=\frac{e^3}3-\frac13\left[\frac{x^3}3\right]_1^e=\frac{2e^3+1}9$.

(b) Vi får $\int\frac1{1+3x^2}\,dx=\left[\begin{array}{c}u=\sqrt3x\\du=\sqrt3\,dx\end{array}\right]=\frac1{\sqrt3}\int\frac1{1+u^2}\,du=\frac1{\sqrt3}\arctan(u)+C=\frac1{\sqrt3}\arctan(\sqrt3x)+C$, så
$$\begin{aligned}\int_1^\infty\frac1{1+3x^2}\,dx&=\lim_{N\to\infty}\int_1^N\frac1{1+3x^2}\,dx=\lim_{N\to\infty}\left[\frac1{\sqrt3}\arctan(\sqrt3x)\right]_1^N\\&=\lim_{N\to\infty}\frac1{\sqrt3}(\arctan(\sqrt3N)-\arctan(\sqrt3))=\frac1{\sqrt3}\left(\frac\pi2-\frac\pi3\right)=\frac\pi{6\sqrt3}.\end{aligned}$$