Answer
$(\ln 4) i+(ln 4) j+(\ln 2)k$
Work Step by Step
We need to integrate the given integral as follows:
$[\int_{1}^{4} (\dfrac{1}{t} dt)]i+[\int_{1}^{4} (\dfrac{1}{5-t} dt)]j+[\int_{1}^{4} (\dfrac{1}{2t} dt)]k=(\ln t)_{1}^{4}i+(-\ln (5-t)]_{1}^{4}k+(\ln \sqrt t)_{1}^{4}k$
or,
$(\ln t)_{1}^{4}i+(-\ln (5-t)]_{1}^{4}k+(\ln \sqrt t)_{1}^{4}k=(\ln 4-\ln 1)i+(-\ln (5-3) -(-(-5-1))]j+\ln 2k$
Thus,
$(\ln 4-\ln 1)i+(-\ln (5-3) -(-(-5-1))]j+\ln 2k=(\ln 4) i+(ln 4) j+(\ln 2)k$
