### Video Transcript

Calculate the integral of one divided by one plus π‘ squared π’ plus one divided by the square root of π‘ squared minus one π£ plus one divided by the square root of one plus π‘ squared π€ with respect to π‘.

We can see weβre asked to calculate an integral. And we can notice something interesting about this Integral; it contains vectors. We notice it contains the unit directional vectors π’, π£, and π€. And we know that thereβs a lot of different notations for vectors. For example, sometimes vectors are represented in bold letters with an underline or with half-arrow notation. And in some instances, unit vectors are given the hat notation. It doesnβt matter which vector notation you prefer. They all mean roughly the same thing. Itβs all up to personal preference.

So in this question, we can see weβre asked to evaluate the indefinite integral of a vector-valued function. And to do this, we need to remember that to evaluate the indefinite integral of a vector-valued function, we just do this component-wise. In other words, weβre just going to evaluate each of our component functions separately. Letβs start with our first component function. We need to find the integral of one divided by one plus π‘ squared with respect to π‘. And to evaluate this, we need to recall one of our rules for differentiating inverse trigonometric functions.

We know the derivative of the inverse tan of π‘ with respect to π‘ is equal to one divided by one plus π‘ squared. But then this means the inverse tan of π‘ is an antiderivative of one divided by one plus π‘ squared. And we can see this is our integrand. So because the inverse tan of π‘ is an antiderivative of our integrand, we can evaluate this integral as the inverse tan of π‘ plus a constant of integration. Weβll call this π.

Now that weβve evaluated the indefinite integral of our first component function, we can now do the same with our second component function. Thatβs the integral of one divided by the square root of π‘ squared minus one with respect to π‘. This time, to help us evaluate our integral, we need to recall one of our rules for differentiating the inverse hyperbolic trigonometric functions. We know the derivative of the inverse cosh of π‘ with respect to π‘ is equal to one divided by the square root of π‘ squared minus one. And once again, this means the inverse cosh of π‘ is an antiderivative of one divided by the square root of π‘ squared minus one, which is our integrand.

So just as we did before, we can evaluate this integral as the inverse cosh of π‘ plus a constant of integration; weβll call this π. Finally, all we need to do is evaluate the integral of our third component function. Thatβs the integral of one divided by the square root of one plus π‘ squared with respect to π‘. And weβll do this in a very similar way to the previous two examples. We need to recall the derivative of the inverse sinh of π‘ with respect to π‘ is equal to one divided by the square root of one plus π‘ squared. So, just like in our previous two examples, this tells us the inverse sinh of π‘ is an antiderivative of our integrand. This means we can evaluate our integral as the inverse sinh of π‘ plus a constant of integration weβll call lowercase π.

Now that weβve found the indefinite integral of all three of our component functions, weβre ready to find the indefinite integral of our vector-valued function. First, the indefinite integral of our first component function was the inverse tan of π‘ plus π. So, in our answer weβll have the inverse tan of π‘ plus π π’. Next, we found the indefinite integral of our second component function to be the inverse cosh of π‘ plus π. So in our answer, weβll have the inverse cosh of π‘ plus π π£. Finally, we found the indefinite integral of our third component function to be the inverse sinh of π‘ plus π. So in our answer, weβll have the inverse sinh of π‘ plus π π€.

And we could leave our answer like this; however, thereβs one more piece of simplification we can do. What would happen if we were to expand our parentheses? If we were to do this, we would end up with the following three terms: π π’, π π£, and π π€.

And remember, π, π, and π are all constants. So π π’, π π£, and π π€ are all constant vectors. This means we could combine all three of these constant vectors into a new constant vector weβll call capital π. And doing this gives us our final answer. Therefore, we were able to calculate the indefinite integral of the vector-valued function given to us in the question by evaluating each integral component-wise. We got the inverse tan of π‘ π’ plus the inverse cosh of π‘ π£ plus the inverse sinh of π‘ π€ plus our constant vector π.