Answer
$r(t)=(\frac{t^2}{2}+1)i+e^tj+(te^t-e^t +2)k$
or $r(t)=\lt \frac{t^2}{2}+1,e^t,te^t-e^t +2 \gt$
Work Step by Step
In order to evaluate the integral we will have to integrate each component of the function individually.
Let $I=\int (ti+e^tj+te^t k)dt$
Thus,
$I=Ai+ Bj+Ck$ ... (1)
Here, $A=t$ and $B=e^t, C=te^t $
Consider $A=t$
Thus, $A=2t=\frac{1}{2}t^2+C$
Now, consider $B=e^t=e^t+C'$
and $C=te^t=te^t-\int e^t(1) dt=e^t(t-1) C''$
Plug in the values of A,B and C in equation (1) and we get
$I=(\frac{1}{2}t^2+C)i+(e^t+C')j+(e^t(t-1) +C'')k$
Here, C,C',C'' are the constants of integration.
We are given $r(0)=i+j+k$, which can be written as $\lt 1,1,1 \gt$.
$r(0)=(\frac{1}{2}t^2+C)i+(e^t+C')j+(e^t(t-1)+ C'')k=j-k$
Hence,
$r(t)=(\frac{t^2}{2}+1)i+e^tj+(te^t-e^t +2)k$
or $r(t)=\lt \frac{t^2}{2}+1,e^t,te^t-e^t +2 \gt$
