91Ó°ÊÓ

First, we need to separate the components of the given vector function \(\mathbf{r}(t)\). $$\mathbf{r}(t) = 2^t\mathbf{i} + \frac{1}{1+2t}\mathbf{j} + \ln(t)\mathbf{k}$$ The components are: $$r_x(t)=2^t$$ $$r_y(t)=\frac{1}{1+2t}$$ $$r_z(t)=\ln(t)$$

Now, compute the indefinite integral of each component individually using integration rules. 1) \(r_x(t)\): $$\int 2^t dt$$ 2) \(r_y(t)\): $$\int\frac{1}{1+2t}dt$$ 3) \(r_z(t)\): $$\int\ln(t)dt$$

Evaluate each integral using the appropriate integration technique. 1) \(r_x(t)\): The integral of an exponential function \(2^t\) can be directly computed as: $$\int 2^t dt = \frac{2^t}{\ln(2)} + C_1$$ 2) \(r_y(t)\): This integral can be computed using a substitution method. Let \(u = 1+2t\) and \(du = 2dt\). Then the integral becomes: $$\frac{1}{2}\int\frac{1}{u}du$$ Now, integrate with respect to \(u\): $$\frac{1}{2}\ln|u| + C_2 = \frac{1}{2}\ln|(1+2t)| + C_2$$ 3) \(r_z(t)\): The integral of \(\ln(t)\) is found using integration by parts: Let \(u = \ln(t)\) and \(dv = dt\). Then, \(du = \frac{1}{t}dt\) and \(v = t\). Using the integration by parts formula: $$\int\ln(t)dt = t\ln(t) - \int t\cdot\frac{1}{t}dt = t\ln(t) - \int 1 dt = t\ln(t) - t + C_3$$

Finally, combine the results of the three indefinite integrals into a new vector function: $$\int \mathbf{r}(t)dt = \left(\frac{2^t}{\ln(2)} + C_1\right)\mathbf{i} + \left(\frac{1}{2}\ln|(1+2t)| + C_2\right)\mathbf{j} + \left(t\ln(t) - t + C_3\right)\mathbf{k}$$ So, the definite integral of the given function is: $$\mathbf{R}(t) = \left(\frac{2^t}{\ln(2)} + C_1\right)\mathbf{i} + \left(\frac{1}{2}\ln|(1+2t)| + C_2\right)\mathbf{j} + \left(t\ln(t) - t + C_3\right)\mathbf{k}$$