If you're just integrating a vector-valued function, you just integrate each component:
[tex]\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt[/tex]
[tex]=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k[/tex]
The first integral is trivial since [tex](\tan t)'=\sec^2t[/tex].
The second can be done by substituting [tex]u=t^2-1[/tex]:
[tex]u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C[/tex]
The third can be found by integrating by parts:
[tex]u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t[/tex]
[tex]\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8[/tex]
[tex]\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C[/tex]
