91Ó°ÊÓ

To integrate the first component, \(te^{t}\), we can use integration by parts. We have to choose our functions \(u\) and \(dv\), such that: $$u = t \quad \text{and} \quad dv = e^{t} dt$$ Now we will compute du and v: $$du = dt \quad \text{and} \quad v = \int e^{t} dt = e^{t}$$ Using integration by parts, we get: $$\int te^{t} dt = uv - \int v du = te^{t} - \int e^{t} dt$$ Compute the remaining integral: $$te^{t} - \int e^{t} dt = te^{t} - e^{t}$$ So, the indefinite integral of the first component is: $$\int te^{t} dt = te^{t} - e^{t} + C_1$$

To integrate the second component, \(t\sin t^{2}\), we can perform the substitution method. We have to choose the appropriate substitution for our integral: $$u = t^2 \quad \text{and so} \quad \frac{du}{dt} = 2t$$ Rearranging for \(dt\), we get: $$dt = \frac{du}{2t}$$ Substituting into the integral: $$\int t\sin t^{2} dt = \int \frac{1}{2} \sin u \ du$$ Now we integrate: $$\frac{1}{2} \int \sin u \ du = -\frac{1}{2} \cos u$$ We need to replace \(u\) with \(t^2\): $$-\frac{1}{2} \cos t^{2}$$ So, the indefinite integral of the second component is: $$\int t\sin t^{2} dt = -\frac{1}{2} \cos t^{2} + C_2$$

To integrate the third component, \(-\frac{2t}{\sqrt{t^{2}+4}}\), we can perform another substitution method. We have to choose the appropriate substitution for our integral: $$u = t^2 + 4 \quad \text{and so} \quad \frac{du}{dt} = 2t$$ Rearranging for \(dt\), we get: $$dt = \frac{du}{2t}$$ Substituting into the integral: $$\int -\frac{2t}{\sqrt{t^{2}+4}} dt = -\int \frac{1}{\sqrt{u}} du$$ Now we integrate: $$-\int \frac{1}{\sqrt{u}} du = -2\sqrt{u}$$ We need to replace \(u\) with \(t^2 + 4\): $$-2\sqrt{t^{2}+4}$$ So, the indefinite integral of the third component is: $$\int -\frac{2t}{\sqrt{t^{2}+4}} dt = -2\sqrt{t^{2}+4} + C_3$$ Finally, the indefinite integral of the given function \(\mathbf{r}(t)\) is: $$\int \mathbf{r}(t) dt = \left( te^{t} - e^{t} + C_1 \right) \mathbf{i} + \left( -\frac{1}{2} \cos t^{2} + C_2 \right) \mathbf{j} + \left( -2\sqrt{t^{2}+4} + C_3 \right) \mathbf{k}$$