Answer
$3 (\ln 3-1)i+(3-e)j+[\ln 3 (\ln (\ln 3)-1)+1] k$
Work Step by Step
We need to integrate the given integral as follows:
$[\int_{1}^{\ln 3} t e^{t} dt]i+[\int_{1}^{\ln 3} e^{t} dt]j+[\int_{1}^{\ln 3} \ln t dt]k=[\int_{1}^{\ln 3} t e^{t}+e^t dt-\int_{1}^{\ln 3} e^{t}]i+[\int_{1}^{\ln 3} e^{t} dt]j+[\int_{1}^{\ln 3} \ln t dt]k$
or,
$[\int_{1}^{\ln 3} t e^{t}+e^t dt-\int_{1}^{\ln 3} e^{t}]i+[\int_{1}^{\ln 3} e^{t} dt]j+[\int_{1}^{\ln 3} \ln t dt]k=[(t-1)e^t]_{1}^{\ln 3}i+[e^t]_{1}^{\ln 3}j+[t\ln t-1]_1^{\ln 3}k$
Thus,
$[3 (\ln 3-1)-(0)e^{0}]i+(3-e) j+(\ln (\ln 3)-1)-(1)(0-1)k=3 (\ln 3-1)i+(3-e)j+[\ln 3 (\ln (\ln 3)-1)+1] k$
