Uppgift 3

(a) Med upprepad partialintegrering får vi
$$\begin{aligned}\int x^3e^x\,dx&=x^3e^x-\int3x^2e^x\,dx=x^3e^x-3\left(x^2e^x-\int2xe^x\,dx\right)\\&=(x^3-3x^2)e^x+6\int xe^x\,dx=(x^3-3x^2)e^x+6\left(xe^x-\int e^x\,dx\right)\\&=(x^3-3x^2+6x-6)e^x+C.\end{aligned}$$
(b) Den primitiva funktionen är
$$\begin{aligned}\int\frac{e^x}{1+e^{2x}}\,dx&=\left[\begin{array}{c}u=e^x\\du=e^x\,dx\end{array}\right]=\int\frac1{1+u^2}\,du\\&=\arctan(u)+C=\arctan(e^x)+C\end{aligned}$$
så vi får
$$\begin{aligned}\int_0^\infty\frac{e^x}{1+e^{2x}}\,dx&=\lim_{N\to\infty}\int_0^N\frac{e^x}{1+e^{2x}}\,dx=\lim_{N\to\infty}[\arctan(e^x)]_0^N\\&=\lim_{N\to\infty}(\arctan(e^N)-\arctan(1))=\frac\pi2-\frac\pi4=\frac\pi4.\end{aligned}$$