Answer
$\pi \mathrm{j}+(\ln 2)\mathrm{k}$
Work Step by Step
Use the last boxed formula of this section:
$\displaystyle \int_{0}^{1}(\frac{4}{1+t^{2}}\mathrm{j}+\frac{2t}{1+t^{2}}\mathrm{k})dt=(\int_{0}^{1}\frac{4}{1+t^{2}}dt)\mathrm{j}+(\int_{0}^{1}\frac{2t}{1+t^{2}}dt)\mathrm{k}$
... the first integral is found in tables .... arctan(t)+C
... the second integral has form $\displaystyle \frac{du}{u}$, where $u=1+t^{2},\quad(du=2tdt)$
$=[4\tan^{-1}t]_{0}^{1}\mathrm{j}+[\ln(1+t^{2})]_{0}^{1}\mathrm{k}$
$=[4\tan^{-1}1-4\tan^{-1}0]\mathrm{j} +[\ln 2 +\ln 1]\mathrm{k}$
$=4(\displaystyle \frac{\pi}{4}-0)\mathrm{j}+(\ln 2-0)\mathrm{k}$
$=\pi \mathrm{j}+\ln 2\mathrm{k}$
