Uppgift 3

(a) Med partialintegrering får vi $\int x^4\ln\lvert x\rvert\,dx=\frac{x^5}5\ln\lvert x\rvert-\int\frac{x^5}5\frac1x\,dx=\frac{x^5}5\ln\lvert x\rvert-\frac15\int x^4\,dx=\frac{x^5}5\ln\lvert x\rvert-\frac15\frac{x^5}5+C=\frac{x^5}{25}(5\ln\lvert x\rvert-1)+C$.

(b) Vi får $\int\frac{dx}{x(\ln(x))^3}=\left[\begin{array}{c}u=\ln(x)\\du=\frac{dx}x\end{array}\right]=\int\frac{du}{u^3}=\frac{u^{-2}}{-2}+C=\frac{-1}{2(\ln(x))^2}+C$ så
$$\int_e^\infty\frac{dx}{x(\ln(x))^3}=\lim_{N\to\infty}\int_e^N\frac{dx}{x(\ln(x))^3}=\lim_{N\to\infty}\left(\frac{-1}{2(\ln(N))^2}+\frac12\right)=\frac12.$$