Uppgift 3

(a) Med upprepad partialintegrering får vi
$$\begin{aligned}I=\int e^{3x}\sin(x)\,dx&=e^{3x}(-\cos(x))-\int3e^{3x}(-\cos(x))\,dx\\&=-e^{3x}\cos(x)+3e^{3x}\sin(x)-9\int e^{3x}\sin(x)\,dx\\&=e^{3x}(3\sin(x)-\cos(x))-9I+D,\end{aligned}$$
vilket ger att $I=\frac{e^{3x}}{10}(3\sin(x)-\cos(x))+C$.

(b) Vi får $\int\frac{x^2}{1+x^6}\,dx=\left[\begin{array}{c}u=x^3\\du=3x^2\,dx\end{array}\right]=\frac13\int\frac{du}{1+u^2}=\frac13\arctan(u)+C=\frac13\arctan(x^3)+C$ så
$$\int_0^\infty\frac{x^2}{1+x^6}\,dx=\lim_{N\to\infty}\int_0^N\frac{x^2}{1+x^6}\,dx=\lim_{N\to\infty}\left(\frac{\arctan(N^3)}3\right)=\frac\pi6.$$