Whew, gave this several shots. All attempts are wrong so here's where I'm at
$$F= ti + \frac{1}{8}j+k; C: r(t)=(e^{8t})i+(e^{8t})j+(-5t^2+t)k, -1\le{t}\le{1}$$ $$W=\int^b_a{\vec{f}\cdot\vec{dr}}$$ $$\frac{\vec{dr}}{dt}=(8e^{8t})i+(8e^{8t})j+(-10t+1)k$$ $$\vec{f}=ti+\frac{1}{8}j+k$$ $$\int^1_{-1}<t,\frac{1}{8},1>\cdot<8e^{8t},8e^{8t},(-10t+1)>dt$$ $$\int^1_{-1}8te^{8t}+e^{8t}-10t+1dt$$ $$8\int^1_{-1}te^{8t}dt+\int^1_{-1}e^{8t}dt+\int^1_{-1}-10tdt+\int^1_{-1}dt$$ $$\int{udv}=uv-\int{vdu}+C$$ $$u=t, dv=e^{8t}, du=1, v=\frac{e^{8t}}{8}$$ $$8\left[\frac{te^{8t}}{8}\Bigg|^1_{-1}-\int^1_{-1}{\frac{e^{8t}}{8}dt}\right]$$ $$8\left[\frac{te^{8t}}{8}\Bigg|^1_{-1}-\frac{1}{8}\int^1_{-1}{e^{8t}dt}\right]$$ $$8\left[\frac{te^{8t}}{8}-\frac{e^{8t}}{64}\Bigg|^1_{-1}\right]$$ $$8\left[\frac{te^{8t}}{8}-\frac{e^{8t}}{64}\Bigg|^1_{-1}=\frac{e^{8t}(8t-1)}{64}\Bigg|^1_{-1}\right]$$ $$8\left[\frac{e^{8t}(8t-1)}{64}\Bigg|^1_{-1}\right]+\int^1_{-1}e^{8t}dt+\int^1_{-1}-10tdt+\int^1_{-1}dt$$ $$\left[8\frac{e^{8t}(8t-1)}{64}+\frac{e^{8t}}{8}\right]\Bigg|^1_{-1}+\int^1_{-1}-10tdt+\int^1_{-1}dt$$ $$\left[8\frac{e^{8t}(8t-1)}{64}+\frac{e^{8t}}{8}+\frac{-10t^2}{2}\right]\Bigg|^1_{-1}+\int^1_{-1}dt$$ $$\left[8\frac{e^{8t}(8t-1)}{64}+\frac{e^{8t}}{8}+\frac{-10t^2}{2}+t\right]\Bigg|^1_{-1}$$ $$\left[\frac{e^{8t}(8t-1)}{8}+\frac{e^{8t}}{8}+\frac{-10t^2}{2}+t\right]\Bigg|^1_{-1}$$ $$\left[\frac{e^{8t}(8t-1)+e^{8t}}{8}+\frac{-10t^2}{2}+t\right]\Bigg|^1_{-1}$$ $$\left[\frac{e^{8t}(8t-1)+e^{8t}+4(-10t^2+2t)}{8}\right]\Bigg|^1_{-1}$$
The correct answer is:
$$W=e^8+e^{-8}+2$$
$$8\int^1_{-1}te^{8t}dt+\int^1_{-1}e^{8t}dt+\int^1_{-1}-10tdt+\int^1_{-1}dt\tag 1$$
$$\int_{-1}^{1}te^{8t} = \frac{t}{8}e^{8t} - \frac{1}{8}\int_{-1}^{1}e^{8t}\tag 2$$
Put (2) in (1)
You get $=te^{8t}|_{-1}^{1} -\int_{-1}^{1}e^{8t}dt ++\int^1_{-1}e^{8t}dt -5t^2|_{-1}^{1} + 2$
Thus you get $$e^{8} + e^{-8}+2$$
The second term in the integration by parts cancels out with the second term of the overall integral and simplifies the process