Question

Compute the indefinite integral of the following functions. $$\mathbf{r}(t)=2^{t} \mathbf{i}+\frac{1}{1+2 t} \mathbf{j}+\ln t \mathbf{k}$$

Short answer

The indefinite integral of the vector-valued function r(t) with respect to t is: $$\int \mathbf{r}(t) dt = \left(\frac{2^t}{\ln 2} + C_i\right)\mathbf{i} + \left(\frac{1}{2}(\ln|1+2t| + C_j)\right)\mathbf{j} + \left(t \ln t - t + C_k\right)\mathbf{k}$$ where Ci, Cj, and Ck are the constants of integration for the i, j, and k components, respectively.

Step-by-step solution

Find the antiderivative of the i component

To find the antiderivative of the i component (2^t), we can use the standard integration formula for exponential functions: $$\int 2^t dt = \frac{2^t}{\ln 2} + C_i$$ where C_i is the constant of integration for the i component.

Find the antiderivative of the j component

To find the antiderivative of the j component (1/(1+2t)), we can use the substitution method. Let u = 1 + 2t, and du = 2dt. Redefine the integral in terms of u: $$\int \frac{1}{1+2t} dt = \frac{1}{2} \int \frac{1}{u} du$$ Now, integrate with respect to u: $$\frac{1}{2} \int \frac{1}{u} du = \frac{1}{2}(\ln|u| + C_j)$$ Substitute u back in terms of t: $$\frac{1}{2}(\ln|1+2t| + C_j)$$

Find the antiderivative of the k component

To find the antiderivative of the k component (ln(t)), we can use integration by parts. Let u = ln(t) and dv = dt, then du = (1/t) dt and v = t. $$\int \ln t dt = t \ln t - \int t \frac{1}{t} dt$$ $$t \ln t - \int dt = t \ln t - t + C_k$$

Combine the components

Now, we can combine the antiderivatives to form the integral of the vector-valued function: $$\int \mathbf{r}(t) dt = \left(\frac{2^t}{\ln 2} + C_i\right)\mathbf{i} + \left(\frac{1}{2}(\ln|1+2t| + C_j)\right)\mathbf{j} + \left(t \ln t - t + C_k\right)\mathbf{k}$$