Uppgift 3

(a) Med partialintegrering får vi
$$\begin{aligned}\int x\cdot2^x\,dx&=\int x\cdot e^{\ln(2)x}\,dx=x\frac{e^{\ln(2)x}}{\ln(2)}-\int\frac{e^{\ln(2)x}}{\ln(2)}\,dx\\&=x\frac{e^{\ln(2)x}}{\ln(2)}-\frac{e^{\ln(2)x}}{(\ln(2))^2}+C=\frac1{(\ln(2))^2}(\ln(2)x-1)2^x+C.\end{aligned}$$
(b) Vi gör en substitution och får:
$$\int_{-\pi/2}^{\pi/2}\frac{\cos(x)}{1+\sin^2(x)}\,dx=\left[\begin{array}{c}u=\sin(x),\\du=\cos(x)\,dx\end{array}\right]=\int_{-1}^1\frac{du}{1+u^2}=[\arctan(u)]_{-1}^1=\frac\pi4-\left(-\frac\pi4\right)=\frac\pi2.$$